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Best JEE Coaching in Itanagar

Maxwell’s equations for electromagnetism in 1887 and Hertz experiments on electromagnetic wave generation and detection strongly supported the wave nature light. Many historic discoveries were made in the same time period, at the end 19th century. These included experiments on the conduction of electricity (or electric discharge) using gases at low pressure in a discharge tub. Important milestones in understanding atomic structure were the discovery of Xrays by Roentgen (1895) and electrons by J. J. Thomson (1897). On applying an electric field to the gas in a discharge tube, at a sufficiently low pressure (0.001 mm mercury column), a discharge occurred between the electrodes. The glass opposite the cathode showed a fluorescent glow. The glass’s colour depended on its type, with soda glass showing a yellowish-green glow. Fluorescence was caused by radiation from the cathode. Best JEE Coaching in Itanagar.

William Crookes discovered these cathode radiations in 1870. He later suggested, in 1879 that they were composed of streams of negatively charged particles. This hypothesis was confirmed by J. J. Thomson (1856-1940), a British physicist. J. J. Thomson applied mutually perpendicular magnetic and electric fields to the discharge tube to determine the speed and specific charge (charge to mass ratio (e/m)) of the cathode-ray particles experimentally. They traveled at speeds of approximately 0.1 to 0.2% above the speed light (3×108 m/s). The current accepted value for e/m at this time is 1.76×1011 C/kg. The value of e/m is independent of the type of metal/material used as the emitter or the gas introduced into the discharge tube. This observation indicated the universality and uniformity of cathode-ray particles. In 1887, it was also discovered that certain metals emit negatively charged particles at small speeds when exposed to ultraviolet light. 

Certain metals also emit negatively charged particles when heated at high temperatures. These particles had a value of e/m that was the same as for cathode-ray particles. These observations proved that these particles were all identical, even though they were produced in different environments. These particles were named electrons by J. J. Thomson in 1897. He suggested that they were fundamental and universal components of matter. He was awarded the Nobel Prize in Physics in Physics in 1906 for his groundbreaking discovery of electrons through his experimental and theoretical investigations into conduction of electricity using gasses. Best JEE Coaching in Itanagar. R. A. Millikan, an American physicist (1868-1953), performed the first oil-drop experiment to measure the electron’s charge. Millikan discovered that the oil-droplet’s charge was always an integral multiple for an elementary charge, 1.602×10 -19 C. 

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Millikans experiment proved that an electric charge can be quantified. The mass (m) for an electron can be determined from the specific charge (e/m) and charge (e/m). 11.2 ELECTRON EMISSION It is known that metals contain free electrons (negatively-charged particles) which are responsible for their conductivity. The metal’s surface is inhospitable to free electrons. An electron trying to escape from the metal surface will be pulled back by the metal’s positive charge. The ions attract the electron to the surface of the metal, holding it there. The attractive pull of the ions means that the electron cannot be released from the metal surface unless it has enough energy. An electron must have a certain amount of energy to pull itself from the metal’s surface. The work function of the material is the minimum amount of energy an electron needs to escape from the surface. It is usually denoted by the letter ph0, and measured in electron volts (electronvolt). Best JEE Coaching in Itanagar.

One electron volt refers to the amount of energy that an electron gains when it is accelerated by a potential differential of 1 volt. 1 eV = 1.6602 x10-19 J. This unit of energy is used commonly in atomic and nuclear physics. The metal’s properties and its surface nature determine the work function (ph0). Table 11.1 shows the work function values for some metals. These values are not exact as they are sensitive to surface impurities. You can see in Table 11.1 that platinum has the highest work function (ph0 = 5.65eV), while caesium has the lowest (ph0 = 2.14eV). Any of the following physical processes can supply the minimum energy necessary to emit electrons from the metal surface:(i) Thermionic emissions: By heating the material to a sufficient temperature, enough thermal energy can be given to the electrons so that they can come out of the iron. The application of an extremely strong electric field of around 108 Vm-1 to a metal can pull electrons out of it.

This is similar to a spark plug. (iii). Photo-electric emission. When light at a suitable frequency is reflected onto a metal surface electrons are released. These photo(light)-generated electrons are called photoelectrons. Heinrich Hertz’s observations Heinrich Hertz (1857-1894) discovered the phenomenon of photoelectric emissions during his electromagnetic wave experiments in 1887. Hertz’s experiments on producing electromagnetic waves using spark discharges revealed that high-voltage sparks were produced across the detector loop when the emitter plate was illuminated with ultraviolet light from an arc lamp. The escape of charged particles from the metal surface was possible by the light shining onto it. These are now called electrons. Some electrons on metal surfaces absorb enough radiation from incident light to overcome the attraction of positive ions. Best JEE Coaching in Itanagar.

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The incident light gives the electrons enough energy to escape the metal’s surface into the surrounding space. 11.3.2 Lenard and Hallwachs’ observations Wilhelm Hallwachs, Philipp Lenard and others studied the phenomenon of photoelectric emissions in detail between 1886-1902. Lenard (1862-1947), who observed that ultraviolet radiations could fall on the emitter plates of an evacuated-glass tube containing two electrodes (metal plate), current flowed in the circuit (Fig. 11.1). The current flow stopped as soon as the ultraviolet radiations stopped. These observations show that ultraviolet radiations falling on emitter plate C causes electrons to be ejected and attracted by the electric field towards positive collector plate A. The current flows because electrons flow through the evacuated tube. Best JEE Coaching in Itanagar.The current flow is caused by light hitting the emitter’s surface. Hallwachs, Lenard and others studied the effects of collector plate potential and incident light intensity on photo current. Hallwachs continued the research in 1888 and connected a negatively charged plate of zinc to an electronscope. 

Hallwachs observed that the zinc plat lost its charge after it was exposed to ultraviolet light. The uncharged zinc plates became positively charged after being exposed to ultraviolet light. The positive charge of a positively charged zinc plat was further enhanced by the use of ultraviolet light. He concluded that the ultraviolet light emitted negatively charged particles from the zinc plate. It was discovered that electrons are emitted from emitter plates by incident light. The electric field pushes the electrons towards the collector plate due to the negative charge. Hallwachs and Lenard observed that no electrons were emitted when ultraviolet light falls on the emitter plates when the frequency of the incident sunlight was lower than a threshold frequency. The nature of the emitter plates determines the frequency at which electrons are emitted. Certain metals, such as zinc, cadmium and magnesium, respond only to ultraviolet light with a short wavelength to emit electrons from the surface. Best JEE Coaching in Itanagar.

Some alkali metals, such as lithium and sodium, potassium, caesium, caesium, rubidium, were sensitive to visible light. When illuminated by light, all these photosensitive substances emit electrons. These electrons were later called photoelectrons after their discovery. This phenomenon is known as the photoelectric effect. This is a schematic representation of the arrangement that was used to study the photoelectric effect. It is made up of an evacuated glass/quartz tub with a photosensitive plaque C and another plate A. Monochromatic radiation from the source S of sufficient short wavelength passes through the W window and falls on C (emitter). A transparent quartz window is attached to the tube. This allows ultraviolet radiation to pass through the tube and irradiate C. The potential difference between plates C and A is maintained by the battery, which can be changed. A commutator can reverse the polarity between plates C and B. The commutator can reverse the polarity of the plates C and A. 

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Thus, electrons will be attracted to the positive collector plate A if it is positive relative to emitter C. The circuit is subject to electric current flow due to the emission of electrons. A voltmeter (V), which measures the potential difference between collector and emitter plates, is used to measure the photocurrent flowing through the circuit. A microammeter (uA) is used to measure the photo current. You can increase or decrease the photoelectric current by changing the potential of the collector plate A relative to the emitter plates C. Also, you can vary the potential difference V between collector A and emitter C. This arrangement is shown in Fig. 11.4.1 Effect of intensity of light on photocurrent 11.4.1 The effect of the intensity of light on photocurrent. Collector A is kept at a positive potential relative to emitter C, so that electrons ejected by C are attracted to collector A. The photoelectric current is measured by changing the intensity of the light. As shown in Figure. 11.2, it is evident that the photocurrent increases linearly as incident light intensities are increased. Best JEE Coaching in Itanagar.

Accordingly, the photocurrent is directly proportional with the intensity of incident radiation. 11.4.2 The potential effect on photoelectric current. We first place the plate A at a positive accelerating potential relative to the plate C. Next, we illuminate the plate C using light with a fixed frequency n (and I1 ). Next, we adjust the plate’s positive potential gradually and measure the photocurrent. The photoelectric current increases as the accelerating potential (positive) increases. The photoelectric current reaches its maximum level when all the electrons emitted by plate A are captured by the plate. This is called saturation. The photocurrent doesn’t increase if we increase the plate A’s accelerating potential. The maximum photoelectric current value is known as saturation current. The case in which all photoelectrons from the emitter plate C reach collector plate A is called saturation current. Now, we apply a negative potential (or retarding) to the plate A relative to the plate C and gradually make it more negative. 

Experimental arrangement is used to study the photoelectric effect. The electrons are repelled when the polarity of the collector is reversed. Only the most energetic electrons can reach the collector A. At a critical, sharply defined value of the plate’s negative potential V0, the photocurrent drops rapidly to zero. The cut-off, or stopping potential, is the value of the negative potential V0 that the plate A must have to be given for a specific frequency of incident radiation. It is easy to interpret the observation in terms photoelectrons. The photoelectrons that are emitted from metal have different energy. The stopping potential must be sufficient to repel all photoelectrons with maximum kinetic energy (Kmax), so Kmax = eV0 (11.1). We can now reproduce this experiment using incident radiation with the same frequency, but with higher intensity (I3 >I2 > I1). The saturation currents have been found to be higher. Best JEE Coaching in Itanagar. This indicates that electrons are being emitted at a higher rate per second in proportion to the intensity of incident radio radiation. The stopping potential for incident radiation of intensity I1 remains the same.

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The maximum kinetic energy for photoelectrons is dependent on the intensity of incident radiation and the light source. 11.4.3 The effect of incident radiation frequency on the stopping potential. We are now studying the relationship between V0 and V0, the frequency of incident radiation. We adjust the intensity of incident radiation to different frequencies, and then study the variation in photocurrent with collector-plate potential. Fig. 11.4. 11.4. The frequency of incident radiations affects the energy of the emitted particles. Higher frequencies of incident radiation have a lower stopping potential. NOTE FROM FIGURE 11.3 Variation in photocurrent with collector potential for different intensities of incident radiation. FIGURE 11.4 Variation in photoelectric current with collector potential for different frequencies. This implies that the maximum kinetic energy of photoelectrons is greater for higher incident radiation frequencies. To stop them completely, we require greater retarding power. There is a minimum cut-off frequency, n 0, for which the stopping power is zero. Best JEE Coaching in Itanagar.

These observations have two implications. (i) The maximum energy of photoelectrons is dependent on the intensity of incident radiation but it varies linearly with frequency. (ii) No photoelectric emission can occur at a frequency n that is lower than the cutoff frequency n0, even though the intensity may be high. The threshold frequency is the minimum cut-off frequency, n0. Different metals have different threshold frequencies. Different photosensitive materials react differently to light. Selenium is more sensitive to light than copper or zinc. Different wavelengths of light can cause different responses from the same photosensitive substance. Photoelectric effects in copper can be triggered by ultraviolet light, but not green or red light. In all of the above experiments, we found that photoelectric emission occurs instantly if incident radiation frequency exceeds the threshold frequency. This is true even when the incident radiation is very dim. Now it is known that emission begins in 10-9 seconds or less. 

This section summarizes the experimental features and observations. (i) The photoelectric current for a given photosensitive matter and frequency of incident radiation (above threshold frequency), is directly proportional with the intensity of incident sunlight (Fig. 11.2). (iii) The saturation current for a photosensitive material with incident radiation frequency is proportional to its intensity, whereas the stopping power is independent of the intensity. (Fig. 11.3). (iii). A photosensitive material has a minimum cut-off frequency for incident radiation. Best JEE Coaching in Itanagar.This frequency is called the threshold frequency. Below this frequency, photoelectrons are not produced, regardless of how intense the incident sunlight. The stopping potential, or equivalently the maximum kinetic FIGURE 111.5 Variation of the stopping potential V0 for a given photosensitive materials with frequency n incident radiation. Dual Nature of Radiation & Matter Energy of Emitted Photoelectrons Increases linearly with frequency, but not independent of intensity (Fig. 11.5). (iv) Photoelectric emission occurs instantly without any time lag (10 to 9s or less), even if the incident radiation is extremely dim.By the end of 19th century, the wave nature of light had been established. 

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The wave picture of light was able to explain the phenomena of interference, difffraction, and polarisation in a natural and satisfying way. This picture shows light as an electromagnetic wave composed of electric and magnet fields that distribute energy continuously over the area of space in which it is extended. Let’s now examine if the wave picture of light can be used to explain the photoelectric emission observations made in the previous section. Wave picture of light shows that the radiant energy is absorbed by the free electrons on the surface of the metal, which is the point at which the radiation beam falls. The amplitude of magnetic and electric fields is proportional to the intensity of radiation. Therefore, each electron should absorb more energy for a given intensity. As you can see, the maximum kinetic energies of photoelectrons at the surface will increase with increasing intensity. No matter the frequency of radiation, an intense beam of radiation should be strong enough to give enough energy to electrons to exceed the energy required to escape the metal surface. Best JEE Coaching in Itanagar.

Therefore, a threshold frequency should not be established. These expectations of wave theory are directly contrary to observations (i), ii, and (iii), which were given at the end sub-section 11.4.3. We should also note that the wave picture shows the electrons absorbing energy continuously across the whole wavefront. Because electrons absorb energy in large numbers, the energy per electron per unit of time is small. According to explicit calculations, it could take several hours for one electron to absorb enough energy to overcome its work function and make it out of the metal. This is in stark contrast to observation (iv), which shows that photoelectric emission occurs instantaneously. The wave picture cannot explain photoelectric emission’s most basic features. 11.6 EINSTEIN’S PHOTOELECTRIC EQUATION – ENERGY QUANTUM of RADIATION Albert Einstein (1879-1955), proposed a radical new picture of electromagnetic radiation in 1905 to explain the photoelectric effect. This picture shows that photoelectric emission doesn’t occur by constant absorption of radiation energy. 

Radiation energy is made up of discrete units, the so-called quanta of radiation energy. Each quantum of radiant heat has an energy hn. Here h is Planck’s constant and the frequency of light. Photoelectric effect is where an electron absorbs some quantum of radiation energy (hn). The electron emits maximum kinetic energy Kmax = Hn – Ph0 (11.2). More tightly bound electrons will have kinetic energies lower than the maximum. The number of photons that pass per second determines the intensity of light at a given frequency. An increase in intensity will result in a greater number of electrons being emitted per second. The energy of each photon determines the maximum kinetic energies of the photoelectrons. This equation explains in an elegant and simple way all of the observations about photoelectric effect made at the end 11.4.3. * Eq. * According to Eq. (11.2), Kmax is linearly dependent on n and independent of the intensity of radiation. Best JEE Coaching in Itanagar. This agreement with observation is also supported by Eq. 

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In Einstein’s image, the photoelectric effect is caused by one quantum of radiation being absorbed by an electron. This basic process does not care about the intensity of radiation, which is proportional to how many energy quanta are per unit area per hour. * Eq. * Eq. The threshold frequencyn0 (=ph0/h), for a metal surface, is below which photoelectric emission cannot be achieved, regardless of how intense or long-lasting the radiation. * The intensity of radiation in this image is proportional to how many energy quanta are available per unit area and per unit of time. For n > 0, the more energy quanta are available, the more electrons absorb the energy quanta, and the higher the number electrons that come out of the metal. For n > 0, the photoelectric current is proportional with intensity. ALBERT EINSTEIN (1879-1855) Albert Einstein (1879-1855) Einstein was born in Ulm (Germany). He is one of the most important physicists ever. He published three groundbreaking papers in 1905. He introduced the concept of light quanta, now called photons, in his first paper. Best JEE Coaching in Itanagar.

This was used to explain the photoelectric effect. He developed a theory for Brownian motion in the second paper. This theory was confirmed experimentally several years later. It provided convincing evidence of the existence of an atomic picture. The special theory of relativity was born from the third paper. He published the general theory on relativity in 1916. Einstein’s greatest later contributions include the idea of stimulated emission, which he introduced in an alternative derivation to Planck’s blackbody radio law, the static model of space that started modern cosmology and quantum statistics for a gas full of massive bosons. He also critically examined the foundations of quantum mechanics. He was awarded the Nobel Prize in Physics in 1921 for his contributions to theoretical physics, and the photoelectric effect. Dual Nature of Radiation & Matter * Einstein’s picture shows that the photoelectric effect is a fundamental elementary process. It involves the absorption by an electron of a light quantum. This is a quick process. 

Photoelectric emission can be instantaneous regardless of the intensity, i.e. the amount of radiation per unit area and time. Because the fundamental elementary process remains the same, low intensity does not necessarily mean delayed emission. The intensity only affects the number of electrons that are able participate in the elementary (absorption by a light quantum electron by one electron) and the photoelectric current. Eq. Eq. (11.1), is the photoelectric equation. (11.1), the photoelectric equation. This is a significant result. Best JEE Coaching in Itanagar.This predicts that V0 versus N curve is a straightline with slope = (h/e), regardless of the nature of the material. Millikan conducted a series experiments on photoelectric effects between 1906 and 1916, which were intended to disprove Einstein’s equation. Millikan measured the slope of the straightline obtained for sodium. Millikan, in 1916, proved Einstein’s photoelectric equation valid, rather than disproving it. In agreement with other experiments, the experimental determinations of h0 and ph0 that were made using the hypothesis light quanta to explain the photoelectric effect led to Einstein’s acceptance of the photoelectric effect picture. 

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Millikan proved the photoelectric equation to be accurate for many alkali metals and a wide range radiation frequencies. 11.7 PARTICLE NATURE of LIGHT: THE POTON The photoelectric effect proved that light in interaction was composed of quantuma, or packets, of energy h n. Does the light quantum of energy have to be associated with a particle or quanta? Einstein reached the crucial conclusion that the light quantum could also be associated (h n/c). The light quantum is associated with a particle if it has a definite value for energy and momentum. Later, this particle was called photon. In 1924, the particle-like behavior of light was confirmed by A.H. Compton’s experiment (1892-1962), which measured scattering of Xrays from electrons. For his contributions to theoretical physics, Einstein received the Nobel Prize in Physics in Physics in 1921. Millikan received the Nobel Prize for Physics in 1923 for his research on the elementary charge and photoelectric effect. Best JEE Coaching in Itanagar.

The photon picture of electromagnetic radiation can be summarized as follows: (i). In the interaction of radiation and matter, radiation behaves like it is made of photons. (iii) Each photon is charged with energy E (=hn), momentum p (=h n/c), speed c (= the speed of light). (iii). All photons that emit light at a given frequency n or wavelength l have the same energy and momentum E (=hn=hc/l) irrespective of the intensity of the radiation. An increase in intensity of light at a wavelength will only result in a greater number of photons crossing an area per second, each photon possessing the same energy. Photon energy does not depend on the intensity of radiation. (iv) Photons have an electrical neutral state and are not affected by magnetic and electric fields. (v) The total energy and momentum of a photon particle collision, such as a photon–electron collision, are conserved. The collision may not preserve the number of photons. A photon could be absorbed, or created from another one. 

WAVE NATURE of Matter The dual (wave-particle and particle) nature of light (electromagnetic radio) is clearly evident from the information in this chapter and in the previous chapters. The phenomena of interference, difffraction, and polarization all show the wave nature of light. In contrast, the Compton effect and photoelectric effect, which involve momentum transfer and energy, radiation behaves like it’s made up of many particles, the photons. The nature of an experiment will determine whether a wave or particle description is the best. Best JEE Coaching in Itanagar.Both descriptions are necessary for understanding the common phenomenon of seeing objects by our eyes. Wave picture describes the gathering and focusing of light by our eye-lens. The photon picture of light describes the absorption of the light by the rods, cones, and retina. The natural question is: Does radiation have a dual (wave-particle?) nature? If so, then might the particles of nature (the protons, electrons, etc.) also have a dual nature? Wave-like characteristics are also found in other particles of nature (electrons, protons, etc.) 

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Louis Victor de Broglie (1892-1987), a French physicist, proposed the bold hypothesis in 1924 that matter particles moving under certain conditions should exhibit wave-like properties. He believed that nature was symmetrical, and that matter and energy must both have symmetrical characteristics. Matter should also have dual aspects if radiation has dual aspects. De Broglie suggested that the wave length of a particle with momentum p can be given as l =h hp m = (11.5), where m is its mass and v is its speed. The de Broglie relation is also known as Equation 11.5. The wavelength l of the matter-wave is called de Broglie wavelength. The de Broglie relationship reveals the dual aspect of matter.The wavelength is so tiny that it’s beyond measurement. This is why the macroscopic objects we encounter in our daily lives do not display waves-like characteristics. However in the sub-atomic realm the character of wave particles is evident and quantifiable. Take one electron (mass m charged e) which is at rest and accelerated by the voltage V. Best JEE Coaching in Itanagar.

The K kinetic energy of the electron is equal to the amount of work completed (eV ) through the electric field. K = eV (11.8) Then K = 1 2m v2 = 2 p m , which means it is that the equation p = M K m eV equals (11.9) The De Broglie wavelength of electrons is 2 2 h the h h pm K m EV (11.10) In addition to the numerical values for the h, m and e and e, we obtain 1.227 Nm V l = (11.11) where V is the value of the accelerating potential in the volts. For a 120 V acceleration potential Equation. (11.11) yields the formula l = 0.112 millimeters. The wavelength is in similar magnitude to the space between the two atomic planes of crystals. This indicates that matter waves that are caused by electrons can be tested using crystal diffraction experiments similar to X-ray diffracted diffraction. We will discuss the proof of concept that supports the de Broglie hypothesis in the following section. It was 1929 when de Broglie was awarded the Nobel Prize in Physics for his work in determining the nature of electrons as waves. 

The matter-wave theory elegantly implemented Heisenberg’s uncertainty rule. Based on the principle, it’s impossible to determine the motion and the position for one electron (or any other particle) simultaneously precisely. Best JEE Coaching in Itanagar.There always exists an element of uncertain ( in the form x) in the description of position , and a degree of uncertainties (p ) when it comes to the definition of momentum. This is because the product between x and is of the form h* (with the formula h = h/2), i.e., the formula x p (11.12) (11.12) Formula (11.12) permits one to speculate that x could be zero, but the p value must be infinite to ensure for the result to be not non-zero. In the same way, if there is no p, it is necessary for x to be infinite. In general, both x and p are non-zero , so they have a product on the form of h. If an electron is able to have a definitive momentum p (i.e.p = 0.) according to the de Broglie relationship the electron has a specific wavelength of l. A wave that is definitive (single). 

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LOUIS VICTOR DE BROGLIE (1892 – 1908 -) Louis Victor de Broglie (1892 – 1987) French physicist who put forward the revolutionary concept of the waves in matter. The idea was created through Erwin Schrodinger into a fullfledged quantum mechanics theory popularly referred to in the field of “wave mechanics. The year 1929 was the time he received with the Nobel Prize in Physics for his research into the electron’s wave nature. A more thorough analysis results in x p= the ratio h/2. (c) NCERT should not be published again 401 Dual Nature and Nature of Matter Wavelength extends across space. According to Born’s probabilistic interpretation, this implies that the electron isn’t restricted to a finite area of space. This means that its location uncertainness is infinity (x + ) and is in line in accordance with the concept of uncertainty. In general the matter wave that is that is associated with the electron doesn’t extend throughout the space. It is a packet of waves that extends over a finite area of space. In this case, x does not have an infinite value however it does have a finite amount determined by the extension in the wavelength of the signal. Best JEE Coaching in Itanagar.

Additionally, it is important to realize that a wave-packet with finite length does not possess a single wavelength. It is composed of wavelengths that are spread out around a central wavelength. Based on de Broglie’s equation then, the energy of the electron is also likely to be spread out – the uncertainty is p. This is expected from the principle of uncertainty. It can be seen that the wave packet explanation along with the de Broglie equation and the Born probability interpretation replicate the Heisenberg uncertainty principle precisely. In Chapter 12 The de Broglie relation will be shown to support Bohr’s postulate on quantisation of the angular force of electrons in an atom. Figure 11.6 depicts a illustration of (a) an isolated wave packet as well as (b) the extended waves that has a fixed wavelength. Example 11.4 What is the de Broglie wavelength that is associated to (a) one electron that is moving at 5.4×106 milliseconds and (b) the ball with a 150 grams in mass traveling at 30.0 milliseconds? Solution (a) for the electron mass m: 9.11×10-31 kg speed v = 5.4×106 milliseconds. 

Then, momentum p = m v = 9.11×10-31 (kg) x 5.4 x 106 (m/s) p = 4.92 x 10-24 kg m/s de Broglie wavelength, l = h/p = . . 34 -24 6 63 10. Js 4 10 kg m/s + l = 0.135 Nm (b) for the ball. Mass is 0.150 kg and speed v’ = 30.0 milliseconds. Then momentum p’ = m’ v’ = 0.150 (kg) x 30.0 (m/s) p’= 4.50 kg m/s de Broglie wavelength l’ = h/p’. Figure 11.6 (a) (a) The wave description of a electron’s packet. Best JEE Coaching in Itanagar.The wave packet is the spread of wavelengths around a central wavelength (and consequently, by de Broglie’s relationship, a spread in momentum). Therefore, it is associated with uncertainty in the position (x) as well as an uncertain acceleration (p). (b) A matter wave that corresponds to a certain energy of an electron is able to extend across the space. In this example there is a p value of zero and x – . Example 11.4 (c) The NCERT is not to be republished as Physics 402 Example 11.6 Example 11.4 + . . 34 6 64 10 Js 4 50 kg m/s x = L 1.47 x 1.47 x10-34m The de Broglie spectrum of electron is similar to the wavelengths of X-rays. 

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But for the ball, it’s around 10-19 times larger than the proton, which is far beyond any measurement made by experiment. Example 11.5 A proton, an electron particle and a proton all have the identical energetic kinetic energy. Which particle has the shortest de-Broglie wavelength? Solution for a particle de Broglie wavelength is L = h/p Kinetic Energy K = p 2 2. Then, l = mK/2 for identical kinetic energy K the de-Broglie wavelength associated to the particles is dependent on the square root of their mass. Protons ( ) 1 1H has a mass of 1836. more massive than an electron, and the particle ( ) 4 2 He is four times the mass of proton. Therefore particles have the most compact de Broglie wavelength. Example 11.5 PROBABILITY INTERPRETATION OF Matter Waves It’s worthwhile to take a moment here to consider what a matter-wave associated with a particle, for instance electrons, for instance, mean. In reality, a complete knowledge of physical properties of radiation and matter is not yet available. Best JEE Coaching in Itanagar.

The pioneers of quantum mechanics (Niels Bohr, Albert Einstein, and numerous others) had to grapple with this issue and other related ideas for quite a while. The deep understanding of the physical aspects that is quantum mechanics has continued be a subject of ongoing research. However, the notion of matter waves has been mathematically introduced into modern quantum mechanics , with great satisfaction. A significant milestone in this area is in the year Max Born (1882- 1970) proposed a probability interpretation of the matter wave’s amplitude. According to thistheory, it is the intensity (square of the magnitude) of the matter-wave at the point will determine the density probability of the particle in that area. Probability density is the measure of probabilities per volume. Therefore, if A represents the magnitude of the sound at the point that is 2 V is the chance that the particles will be located in a small area V around the area. Therefore, if the magnitude of a matter wave is high within a specific area, it is a higher probability of the particle being discovered there than in areas where the intensity is low. 

 A particle moves three times faster than an electron. Best JEE Coaching in Itanagar. Its ratio between the wavelength of de Broglie of the particle to electron is 1.813 10-4. Determine the mass of the particle and determine the particle’s mass. Solutions de Broglie the wavelength for a moving particle with mass m and speed of v: (c) NCERT not to be published 403 Dual Nature Radiation and Matter h h P mv l = Mass, m = h/lv for an electron weight me equals h/le we have V/V = 3 and L/LE = 1.813 10-4. In the end the it is the mass of the particles me e V e e v LM = (9.11×10-31 kg) (9) x (1/3) (1/1.813 x 10-4) (1/1.813 10-4) the result is 1.675 1027 kg. Therefore, the particle that has this mass, could be a proton or neutron. Example 11.7 What is the wavelength of the de Broglie that is that is associated with an electron which is being accelerated with a potential difference that is 100 V? Solutions Accelerating voltage V=100 V. The wavelength of de Broglie is l = h 1 227/p . V = nm l 1 227 . 100 = nm 0.123 nm The de-Broglie wavelength of an electron in this example is of the order of wavelengths for X-rays. 

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The wave nature of electrons first experimentally confirmed by C.J. Davisson and L.H. Germer in 1927, and later by G.P. Thomson at the time of 1928. They discovered the effects of diffraction when the scattering of electrons by crystals. Davisson and Thomson were awarded with Thomson the Nobel Prize in 1937 for their discovery of the Diffraction of electrons caused by crystals. The arrangement they used for their experiments for the study by Davisson and Germer is schematically illustrated in Fig. 11.7. It is made up in an electronic gun, which includes a tungsten-based filament F, which is coated with barium oxide, and heated by an energy source of low voltage (L.T. or battery). The electrons released by the filament are then accelerated to the desired speed. (or battery). FIGURE 11.7 Best JEE Coaching in Itanagar. Davisson-Germer electron Diffraction arrangement. Example 11.6 Example 11.7 (c) (c) is not to be published again Physics by applying suitable voltage and potential from an energy source with high voltage (H.T. or battery). 

They pass through a cylinder that has tiny orifices along its length. This creates an extremely collimated beam. The beam is then made to hit the surfaces of the crystal made of nickel. Electrons scatter throughout the directions through the molecules in the crystal. Its intensity scattered in a specific direction, is recorded through the detector (collector). The detector is moved in a circular manner and connected to a sensitive galvanometer that records the current. The galvanometer’s deflection corresponds to the strength of electrons that enters the collector. The device is contained within an air-filled chamber. The detector is moved on the circular scale in different places how intense the beam of scattered electrons is measured at different levels of the angle of scattering, (which is the difference between incident beam and scattered beams of electrons. 

The change in the strength (I) of scattered electrons based on the angle of scattering measured for various speeds and voltages. The experiment was carried out by altering the accelarating voltage between 44 V and one of 68 V. It was discovered that there was a significant peak with respect to the magnitude (I) that was scattered by the electrons at an accelarating power of 54V with a scattering angle of 50o. It is believed that the appearance of the peak within a certain direction is caused by the constructive interference of electrons scattered by various layers within the atoms that are regularly spaced in the crystals. Through the measurement of electron diffraction this wavelength was discovered at 0.165 nm. Best JEE Coaching in Itanagar. The wavelength de Broglie that is associated with electrons, using Eq. (11.11) for V = 54V, is calculated as l = h 1 227 V / p = . nm l 1 227 = . 0.167 nm 0.167 nm. Therefore there is a great relationship between this theoretical and measured value of the de Broglie’s wavelength. 

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DavissonGermer experiment therefore demonstrates the nature of electrons as wave-like and the de broglie connection. In 1989, more recently the wave nature of beams of electrons was demonstrated through a double-slit experiment like the one used to demonstrate the nature of light waves. In an experiment in 1994, there were interference fringes created using Iodine molecules’ beams, that are approximately one million times larger than electrons. This de Broglie hypothesis has been essential to the evolution of quantum mechanics in the modern era. It also has resulted in electronic optics as a field. The properties of electrons’ waves have been utilized in the development of an electron microscopes, which are an amazing improvement, and has greater resolution than optics microscope. Development of electron microscope (c) NCERT not to be republished 405 Dual Nature of Radiation and Matter Best JEE Coaching in Itanagar

The energy required by an electron for it to emerge from a surface made of metal is known as”the work function” of the material. The energy (greater than that of the work function (pho ) needed to emit electrons from the surface of the metal can be obtained by heating it to a suitable temperature or applying a strong electrical field or by light with a suitable frequency. 2. Photoelectric effect refers to the process of the emission of electrons by metals that are illuminated by light that is of a frequency suitable to the metal. Certain metals are sensitive to ultraviolet light , while others are even sensitive to visible light. Photoelectric effect is the process of converting the energy of light into electricity. It is governed by the conservation of energy law. Photoelectric emission happens instantly and comes with a variety of special characteristics. 3. The photoelectric current is influenced by (i) how intense the light incident, (ii) the potential difference between the two electrodes as well as (iii) the properties of the material that emits light. 

The potential for stopping (Vo ) is dependent on (i) the frequency of light incident as well as (ii) the type of the material that emits it. If a particular frequency is the light that is incident, it will be independent of the intensity. It is linked to the maximum energy kinetic of the electrons that are released: V0 = (1/2) M v 2 maximum = Kmax . 5. If a frequency is below a certain level (threshold frequency) (n0), which is the characteristic of the metal, there is no photoelectric discharge occurs regardless of how intense the intensity. 6. The traditional wave theory can not explain the primary characteristics of the photoelectric effect. The theory of constant absorption of energy by radiation was not able to explain the dependence from Kmax on intensity and the existence of no and the speed of the process. Einstein explained these phenomena by referring to the photon images of light. Best JEE Coaching in Itanagar.According to Einstein the light spectrum is made up of small, discrete pieces of energy known as photons or quanta. 

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Each photon is a source of energetic energy (= h N) and momentum (p) (is h/l) that are based on its frequency (n ) of light incident and not its intensity. Photoelectric emission that is emitted from the metal surface is caused by the absorption of a photon electrons. 7. Einstein’s equation for photoelectricity is in line to the law of conservation of energy in relation to the absorption of light by electrons in the metal. The maximum energy of the kinetic (1/2)m V 2 Max is the same as the energy of a photon (hn ) less working function (ph0) (= hn zero ) of the metal in question 1 2 m v2 max = V0 + e Hn (hn – ph0) = (n + nu zero ) The photoelectric equation is the basis for understanding all the characteristics in the phenomenon of light. Millikan’s first exact measurements confirmed Einstein’s equation of photoelectricity and determined the precise number of Planck’s constant h . This resulted in the acceptance of the particle or photon descriptions (nature) for electromagnetic radiation as proposed by Einstein. 8. Radiation can be described as a dual particle and wave. Best JEE Coaching in Itanagar.

The nature of the experiment determines which particle description is the best to comprehend the result of the experiment. The idea that radiation and matter must have symmetry in their nature Louis Victor de Broglie (c) NCERT was not to be repeated Physics 406 attributed a wave-like nature in matter (material particle). The waves that are associated with moving particles are referred to as Matter waves, also known as de Broglie waves. 9. The wavelength of de Broglie (l) that is associated with the motion of a particle is connected to its momentum p in the form the equation L = h/p. The dualistic nature of matter is part of the de Broglie relationship, that includes an idea of wave (l) as well as the concept of particle (p). De Broglie’s wavelength can be not dependent on the nature and charge of the particle. It can be measured in a significant way (of the order of distance between the atomic planes of crystals) only in the case of sub-atomic particles such as electrons protons,. (due due to the small size of their masses , and thus, their moments). 

But, it’s extremely small, and far beyond measurement in the case of macro objects that are which are often encountered in daily life. 10. Electron Diffraction experiments conducted performed by Davisson and Germer as well as the work of G. P. Thomson and G. P. Thomson, along with numerous other experiments have confirmed and verified the electron’s wave nature. This de Broglie hypothesis of matter waves supports Bohr idea about stationary orbits. Physical Symbol Dimensions Unit Remarks Quantity Planck’s H [ML2 T 1 ] J s E = constant hn Stopping [ML2 T-3A-1] E V0 is Kmax’s potential. Work Ph0 (ML2 T -2 ] J Kmax eV = E-ph0 Threshold n0 = T -1[T -1] Hz n0 = ph0 De Broglie’s l = h/h frequency [L] m l = HP wavelength POINTS TO CONSIDER 1. The electrons that are free in a metal are free because they are able to move within the metal at an unchanging potential (This is only an approximate). They cannot leave the metal. They require additional energy to break away from the iron. Best JEE Coaching in Itanagar.

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The free electrons that are present in a metal don’t all possess the same amount of energy. As with molecules in a container, electrons possess an energy distribution that is specific to an appropriate temperature. This distribution differs from the typical Maxwell’s distribution you’ve learned about in the study of the kinetic theory of gas. You’ll learn more about it in subsequent courses, however the distinction lies to the fact that electrons follow Pauli’s exclusion rule. 3. Due to the distribution of energy of electrons free in metals, the amount of energy required by an electron in order to be released from the metal varies with respect to the different electrons. Electrons with greater energy need less energy to get through the iron than those that have smaller energies. The work function represents the smallest amount of energy needed by electrons to exit the metal. The observations on the photoelectric effect suggest that, in the case of matterlight interactions the absorption of energy is carried out in distinct elements of the hn. 

This isn’t identical to that of saying that light is composed of particles, with each having the energy the form of hn. 5. The stopping potential is the subject of study. (its independence of its intensity as well as its dependence on frequency) is the most important criterion between the photon-picture and wave-picture of photoelectric effects. 6. A wavelength for a matter wave as measured by the formula h p l = is of no physical significance. Its velocity vp does not have any significance in terms of physical properties. However its group velocity is relevant and is equal to what the speed of the particle.The dimensions of the nucleus are significantly smaller than the dimensions for an atom. Studies on the scattering of particles showed it was the case that the radius of the nucleus is less than the radius of an atom , by a factor of around 100 . This means that the size of a nucleus is around 10-12 times the size of an atom. That is the atom is nearly empty. If an atom were to be expanded enough to fill a schoolroom the nucleus will be the size of a pinhead. Best JEE Coaching in Itanagar.

However, the nucleus is home to the majority (more than 99.9 percent) in the mass and weight of an atom. Does the nucleus possess an atom-like structure, like the atom? If yes What are the constituents in the nucleus? How do they hold them together? What is the mechanism that holds them together? In this section, we will explore the answers to such questions. We will look at various aspects of nuclei, such as their size, mass , and stability, as well as with nuclear phenomena like nuclear fission, radioactivity and nuclear fusion. 13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS The weight of an atom is extremely tiny, when compared to a kilogram. For instance the mass of carbon atoms 12C, for instance, is 1.992647 10-26 kg. Kilogram is not a efficient way to measure these small amounts. So, aNuclei unit is used to describe the mass of atoms. This is the nuclear mass unit (u) that is which is defined as 1/12th of mass of carbon (12C) atom. Best JEE Coaching in Itanagar.The mass atomics of different elements, that are expressed in the the atomic mass units (u) are near in relation to the masses of the hydrogen atom.


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However, there are several remarkable exceptions to this principle. For instance the mass of an atomic the chlorine atom has been measured to be 35.46 u. The precise measurements of the mass of an atom are done using the mass spectrometer. The measurements of the atomic masses reveal the existence of several kinds of atoms from the same element that have similar chemical properties however, they differ in terms of mass. These atomic species of the same element that differ in mass are referred to as isotopes. (In Greek, isotope means the same spot, i.e. they share the same location on the periodic table.) It was found that virtually every element is made up of an assortment of isotopes. The amount of isotopes is different between elements. Best JEE Coaching in Itanagar. Chlorine for instance has two isotopes with masses of 34.98 U and 36.98 u that are almost an integral multiple of the mass of the hydrogen atom. 

The relative abundances of these isotopes are 75.4 and 24.6 per cent, respectivelyDiscovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. However, the masses of the nuclei from tritium, deuterium, and hydrogen are approximately 1:2:3. So, the nuclei of tritium and

deuterium must have as well as the proton, a small amount of neutral matter. There is a lot of neutral material inside the nuclei of the isotopes measured in mass units of proton, is equivalent to one and two in each. This suggests that the nuclei of atoms are filled with as well as proton particles, neutral matter in many different units of a fundamental unit. This theory was confirmed during 1932, by James Chadwick who observed emission of neutral radiation after beryllium nuclei were bombarded by alpha particles. (a-particles constitute helium nuclei to be covered in a future section). It was discovered that this radiation neutral could remove protons from light nuclei , such like those of helium nitrogen and carbon. 

The only known neutral radiation at the at the time were the photons (electromagnetic radiation). Utilizing the concepts of conservation of energy and momentum proved that if the radiation was composed of photons the energies of photons would need to be significantly higher than the energy available from radiation of the beryllium nuclei by particles called a-particles. Best JEE Coaching in Itanagar. The key for this mystery, that Chadwick successfully resolved, was to suppose that neutral radiation consisted of a brand new kind of neutrons, which are neutral particles. Through the conservation in energy as well as momentum Chadwick was able to calculate the mass of this new particle that is’very nearly similar to the mass of a proton’. A neutron’s mass is recognized to a great level of precision. It is m nu equals 1.00866 the u value is 1.6749×10-27 kilograms (13.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron. The neutron free of charge, unlike the proton that is free it is unstable. It degrades into a proton, an electron , and an neutrino (another fundamental particle) as well as an average life of around 1000 years. 

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However, it is solid inside the nucleus. The structure of a nucleus can be described with the following words and symbols: Z – atomic number = the number of proton [13.4(a) N neutron number = the quantity of neutrons [13.4(b) Mass number Z + N = the total number of neutrons and protons [13.4(c)The other employs the term “nucleon” to refer to either a neutron or proton. Therefore, the number of nucleons within an atom is the mass number, A. Nuclides, or nuclear species, are represented by the notation X A Z , where”X” is the symbol for chemical used to describe the species. For instance the nucleus of gold is represented by the number the number 197 779 Au. It has 197 nucleons including 79 of them being protons, and the rest are neutrons. (c) NCERT should not be republished as 441 Nuclei. composition of isotopes in an element is now easily understood. The isotopes of an element have the same amount of protons but differ in the amount of neutrons. Deuterium 2 1 H which is an isotope made of hydrogen, includes one proton as well as one neutron. Best JEE Coaching in Itanagar. Its tritium isotope 3 1 H includes one proton as well as two neutrons. 

The element gold contains 32 isotopes that range from A = 173 to A = A =. As we have mentioned before, the elements’ chemical properties depend from their electron structure. Since the isotopes’ atoms possess the same electronic structure, they share the same chemical properties and are in the same spot within the periodic table. All nuclides having the identical mass numbers A are known as isobars. For instance the nuclides 3 1 H as well as 3 2He, are considered isobars. Nuclides that have the identical neutron numbers N but with different atomic numbers Z, like that 198 80 Hg is 1907 79 Au , are known as isotones. 13.3 SIZE of the nucleus As we’ve seen in Chapter 12 Rutherford was the first to proposed and proved that there is an nucleus atomicum. Based on Rutherford’s suggestions, Geiger and Marsden performed their famous experiment on the scattering of particles of a-particles by the thin foils of gold. Their research showed that the distance of closest contact with a gold nucleus in a-particles that has kinetic energy 5.5 MeV is about 4.0 10-14 meters. 

The scattering caused by a particle due to the gold sheet could be explained by Rutherford by concluding that coulomb repulsive force is the only one the cause of scattering. Because this positive charge limited to the nucleus itself, the dimensions of the nuclear nucleus must to be smaller than 4.0 10-14 meters. If we employ particles with greater energies that are greater than 5.5 MeV, the distance from the nearest contact with the gold nucleus is less and eventually the scattering will be affected by forces of nuclear short-range which diverge from Rutherford’s calculations. Rutherford’s calculations are based upon pure coulomb attraction among the positively charged aparticle as well as the nucleus of gold. From the distances that the deviations are set the nucleus, the size of the nuclear can be calculated. Best JEE Coaching in Itanagar. When conducting scattering tests that use fast electrons instead of particles, are projectiles that pound targets that are composed of various elements, the dimensions of the nuclei from various elements have been precisely determined. 

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It was discovered that a nucleus having mass A is radial in size. of R equals R0 A/3 (13.5) where R0 is 1.2 10-15m. This indicates that the size of the nucleus is proportional to the R3 is proportional to the A. The density of the nucleus is steady, regardless of the A in all the nuclei. Different nuclei are similar to drops of liquid that has a the same density. Nuclear matter has a density 2.3 1017kg m-3. This is quite a lot in comparison to the ordinary matter, for instance water, which has a density of approximately 103 kg m-3. This is natural, since we’ve seen before that the majority of an atoms are empty. The normal matter that is composed of atoms is filled with a huge quantity of empty space.Mass is energy. Einstein proved through the theory of special relativity, that it was important to view mass as a different kind of energy. Prior to the development of this concept of special relativity,, it was thought that energy and mass kept separate during reactions. Top JEE Coaching in Itanagar. But, Einstein showed that mass is an alternative form of energy that is able to convert energy from mass into other types of energy, like the kinetic energy, and vice versa. 

Einstein provided the well-known mass-energy-energy-equivalence relationship E = Mc2 (13.6) The energy equivalent to mass m is calculated by the equation above and c is the speed of light in a vacuum. It is about 3×108. m s-1Nuclei , which only affects the nucleons that are close to it. This is also known as the saturation property of nuclear force. (iii) A very heavy nucleus, for example A of 240 is bound with less energy per nucleon as compared to the nucleus that has A equal to 120. So, if a nuclear the nucleus A = 240 is broken in two A = 120 nuclei, the nucleons become much more closely bound. This means that energy will be released as a result. It is a significant issue for the production of energy through fission. This will be discussed further in section 13.7.1. (iv) Think about two light nuclei (A 10) merging to create an atomic mass that is heavier. It is apparent that the binding power per unit for the heavier nuclei that have been fused is higher that the energy of binding per nucleon for the nuclei with lighter mass. This means that the resulting system is more closely bound than the first system. In addition, energy will be released during a process of Fusion. 

This is the source of energy of the sun, which will be explained in section 13.7.3. 13.5 NUCLEAR FORCE The force that governs the movement of electrons within atomic nuclei is known as the Coulomb force. In Section 13.4 we’ve discovered that for nuclei with average mass the energy required to bind a nucleon is about 8 MeV which is a lot more than the energy of binding in the atoms. So, in order to tie two nuclei, there needs to be an attraction force of a completely different type. It has to be sufficient to overcome the attraction between (positively positively charged) protons, and to bring neutrons and protons into the small nucleus. We’ve already observed that the steadfastness of the binding energy per nucleon could be seen by its short-range. Numerous aspects that make up the force of nuclear binding can be described below. They are derived from a myriad of experiments that were conducted from 1930 to 1950. Top JEE Coaching in Itanagar. (i) It is more powerful that the Coulomb force that is acting between charges, or the gravitational force between masses. The force of nuclear binding has been able to prevail in comparison to the Coulomb repulsive force acting between protons within the nucleus. 

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This happens since the force of nuclear is more powerful than the Coulomb force. It is smaller than even Coulomb force. (ii) the nuclear energy between nucleons decreases quickly to zero when their distance is greater than one femtometer. This causes an oversaturation of forces in a medium or huge nucleus. This is the main reason for the constant nuclear energy binding per. A rough graph of the energy potential between two nucleons as a result of distance is displayed in Fig. 13.2. Potential energy is minimal at a distance that is approximately 0.8 Fm. This implies this force will be appealing for distances that are greater than 0.8 Fm, and is repulsive when they(iii) the nuclear forces between neutron-neutron proton-neutron and proton is about the similar. Top JEE Coaching in Itanagar. The nuclear force doesn’t have to be dependent on electric charges. In contrast to Coulomb’s laws or Newton’s law of gravitation , there is no mathematical formula of nuclear force. 13.6 Radioactivity A. H. Becquerel discovered radioactivity in 1896 completely through accidental. 

In his research into the phosphorescence and fluorescence of compounds exposed to the visible spectrum, Becquerel discovered an intriguing phenomenon. After shining a few pieces of uranium-potassium sulfurate using visible light He wrapped the pieces in black paper and separated the box from a photographic plate using an element of silver. When, after a few hours of exposure and exposure, the photographic plate developing, it began to show darkening caused by something released by the compound. It could be absorbed by both the black paper and the silver. The subsequent experiments showed radioactivity to be an atomic phenomenon where unstable nuclei undergoes decay. This is known as radioactive decomposition. Three kinds of radioactive decay can be observed in nature: (i) A-decay, where a Helium nucleus 4 2He emits; (ii) b-decay in which positrons or electrons (particles that have similar mass to electrons, but having an energy charge that is opposite to electrons) are released; (iii) G-decay where high-energy (hundreds in keV or greater) photons are released. 

Each of these types of decay will be discussed in the following subsections. 13.6.1 The law of decay in radioactive substances in any radioactive substance that undergoes a, b , or decay in g, it is observed that the amount of nuclei that undergo decay in a unit of duration is proportional to number of nuclei present in the sample. When N represents the nuclei present in the sample, and they undergo decay within the time t, then is N t or N/t = lN (13.10) where L is”radioactive decay”, or disintegration constant. Top JEE Coaching in Itanagar.  The variation in the number of nuclei present in the sample is dN = -N in the period t. So the change rate in N (in the range t – 0.) (d – d N). T = l * N is the number of nuclei which change, which it is never negative. The variation in N, which can be expressed in either of two ways. In this instance, it’s negative since, out of the original N nuclei, the nuclei have been destroyed leaving (N-N) Nuclei. (c) These NCERT are separated at smaller than 0.8 fm.NUCLEAR Energy The curve for binding energy per nuclear nucleon, Ebn shown in figure. 13.1 It has a lengthy flat middle region that runs between A equal 30 to A = 170. In this region, the bound energy of a nucleon remains almost always the same (8.0 MeV). 

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For the nuclei with lighter mass, A < 30, and for the heavier nuclei region, A > 170, the binding energy per nucleon is lower than 8.0 MeV, as we have previously noted. The higher it is the energy of binding, the smaller is the mass of a system bound, like the nucleus. Therefore, if nuclei that have less binding energy convert to nuclei with more binding energy that will result in an release of energy. This is the result when a nucleus that is heavy decays into more or less intermediate mass pieces (fission) and when nuclei with light energy combine into a bigger nucleus (fusion.) Exothermic chemical reactions are the basis of the traditional energy sources, such as petroleum or coal. In this case, the energy levels involved are within the range of 13.4 b-decay of the 13.4 b-decay of the 28-60 Ni nucleus, followed by emission of two g-rays due to deexcitation of the daughter nucleus 2860 (c) (c) Ni . Not to be published again. Physics 452 electronvolts. However when you have a nuclear reaction the energy released is on the order of MeV. For the same amount of material, nuclear sources release an energy output that is a million times greater than chemical sources. The fission of one kilogram of uranium for instance produces 10.14J of power. Top JEE Coaching in Itanagar. 

Compare this to burning 1 kg of coke which produces 107 J. 13.7.1 Fission The possibilities for new possibilities are revealed as we explore the limits of natural radioactive decays and investigate nuclear reactions by bombarding nuclei other nuclear particles, such as neutron, proton, a-particle as well as other particles. One of the most significant neutron-induced reaction is the fission. A good example of fission is when a uranium-isotope of 235 92U that is bombarded by neutron fragments into two nuclear fragments with intermediate mass. 1 235 236 144 1 0 92 % 92, 56, 36 N U U BaKr 3 n0 + + (13.26) Top JEE Coaching in Itanagar. The same reaction could produce additional types with intermediate mass pieces. For instance, 1 236 236 133 99 1. 236 133 99 1 n U U Sb 4 n0 + + – + (13.27) As an example 1 235 140 1 0 54 38 n Nb Xe 2 n0 + – + (13.28) The fragments are radioactive nuclear nuclei. They release b particles sequentially in order to create stable products. Energy released (the Quantum value ) during the fission process of nuclei such as uranium is around 200 MeV per fissioning nuclear nucleus. It is calculated by following Let’s consider the nucleus that has A = 240, which breaks into two pieces with A equals 120. 

It is then Ebn in the case of A of 240 is around 7.6 MeV, Ebn for the two A = 120 fragment nuclei is around 8.5 MeV. The gain in binding energy of nucleons is approximately 0.9 MeV. Thus, the increase of binding energy equals 240×0.9 which is 216 MeV. The energy of disintegration in fission reactions first manifests as the kinetic energy of neutrons and fragments. Then it transfers to the surrounding matter , appearing as heat. The energy source in nuclear reactors that produce electricity can be found in nuclear fission. The huge amount of energy released in an atom bomb results from an uncontrolled nuclear fission. We will discuss in the following section about how the nuclear reactor works. 13.7.2 Nuclear reactor Note one important aspect in the fission reactions described in the equations. (13.26) to (13.28). There is the release of an extra neutron (s) during the process of fission. In the average 2 1/2 neutrons are released during the fission process of the nuclear uranium. This is a tiny fraction because during certain fission events, two neutrons are produced by nuclei fission. It has the elements of fuel properly fabricated. The fuel can be an enriched form of by uranium (i.e. the one that is more abundant in 235 922 U than the naturally occurring uranium). Top JEE Coaching in Itanagar.