aiims physics modern physics II 12 eng

CLASS 12th Modern Physics-II Modern Physics-II 1. Thomson’s Atom Model From the study of discharge of electricity th...

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CLASS 12th

Modern Physics-II

Modern Physics-II

1. Thomson’s Atom Model From the study of discharge of electricity through gases, it became clear that an atom consists of positive and negative charges. As the atom is electrically neutral, the number of positive and negative charges (electrons) must be equal. A theoretical explanation for the structure of atom is called an atom model. In 1898, J.J. Thomson gave the first explanation about the arrangement of positive charges and the electrons inside the atom. According to him, an atom is a sphere of positive charges having radius of the order of 10-10m. The positive chare is uniformly distributed over the entire sphere and the electrons are embedded in the sphere of positive charges just like seed in a watermelon or plums in the pudding as shown in Figure. For this reason, Thomson’s atom model is also known as plumpudding model. The total positive charge inside the atom is equal to the total negative charge carried by electrons, so that every atom is electrically neutral. The electrons were supposed to be located in the cloud of positive charges, such that the system is stable*. If the atom gets slightly perturbed, the electrons in the atoms oscillated about their equilibrium position and result in the emission of radiation of definite frequencies in the form of infra-red, visible or ultraviolet light. Failure of Thomson’s atom model. Thomson’s atom model is a plausible model for atom only apparently. It had to be discarded, because of th following limitations: (i) It could not explain the origin of the spectral lines in the form of series as in the case of hydrogen atom. (ii) It could not account for the scattering of -particles through large angles (even upto 180o) as observed in the case of Rutherford’s -scattering experiment.

2. Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom

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Modern Physics-II H. Geiger and E. Marsden performed an experiment on -particles scattering, in 1911, as suggested by Ernst Rutherford.  Bi radioactive In this experiment, they used a beam of 5.5 MeV, -particles obtained from  source and bombarded it on a thin gold foil. Scattering of -particles was observed though a rotatable detector made up of zinc sulphide screen and a microscope. Observations : (i) Most of the -particles passed though the foil without any deviation. (ii) About 0.14% of the incident -particles scattered by more than 1o. (iii) Deflection of more than 900 was observed in about 0.0125% of the incident -particles. Inferences : (i) Most of the space in an atom is unoccupied as about 99.86% -particles passed without deviation. (ii) There must be an extremely small region of concentrated positive charge at the centre of an atom. This small region is called nucleus. The scattering of -particles is due to encounter between the -particle and the nucleus of the atom. (iii) The nucleus of the atom is so massive as compared with the -particles that it remains at rest during the encounter, whereas electrons, owing to their little mass, cannot appreciably deflect the far more massive -particles. (iv) Electrons revolve around the nucleus in orbits just like planets revolve around the sun. (v) When an -particle strikes the metal foil, it can penetrate the outer electron cloud and approaches the nucleus closely. It then moves under the action of coulomb’s force of repulsion, and its path is hyperbola with the nucleus as the external forces.

3. Impact Parameter Let us consider an -particle travelling with velocity  towards a nucleus having charge Z e. In case, the -particle travels along a path as shown in Figure, then after coming close to the nucleus up to certain distance, it scatters along a path making an angle  with the initial path. The scattering of an -particle from the nucleus of an atom depends upon the impact parameter. Impact parameter of the alpha particle is defined as the perpendicular distance of the velocity vector of the alpha particle from the centre of the nucleus, when it is far away from the atom. It is denoted by b. Rutherford derived the relation between impact parameter and the scattering angle, which is as given below :    cot b=         Discussion. Following inferences can be drawn from the above equation: (i) If the impact parameter b is large, then cot  is also large i.e. the angle of scattering  is small and vice-versa. Thus, if an -particle has large impact parameter, it gets scattered though a very small angle and may practically go undeviated and if he -particle has small impact parameter, it will get scattered though a large angle.

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Modern Physics-II (ii)

If the impact parameter b is zero, then  = 90o cot  = 0 or  = 180o or Thus, if an -particle has impact parameter zero (i.e. if the -particle travels directly towards the centre of the nucleus), it will be scattered through 180o

4. Distance of Closest Approach Let r0 be the distance of closest approach of the -particles to the nucleus. The (positive) charge on he nucleus is Ze, and that on the -particles is 2e, where e is the electronic charge. Therefore, the electrostatic potential energy or the particle at the instant of closest approach is               At this instant, the -particles is momentarily at rest and the initial kinetic energy E is entirely converted into electrostatic potential energy. Hence, at this instant      





       

...[Distance of closest approach]

This is the expression for the distance of closest approach r0 of the -particles. It shows that for a given nucleus, r0 depends upon the initial kinetic energy E of the -particle. When the kinetic energy E exceeds a certain value, the distance of closest approach r0 becomes so small that the nucleus no longer appears as a point charge to the -particle. Then the Coulomb’s inverse-square law and hence the Rutherford formula breaks down.

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Modern Physics-II

5. Drawbacks of Rutherford’s Atom Model Though Rutherford’s atom model explained the results of the -scattering experiment very well, it offered serous difficulties as regards the stability of the atom. The drawbacks of Rutherford’s atom model are as discussed below : (i) When the electrons revolve round the nucleus, they are continuously accelerated towards the centre of the nucleus. According to Lorentz, an accelerated charged particle should radiate energy continuously. Therefore, in the atom, a revolving electron should continuously emit energy and hence the radius of its path should go on decreasing and ultimately, it should fall into the nucleus as shown in figure. However, electrons revolve round the nucleus without falling into it. Thus, Rutherford’s atom model cannot explain the stability of the atom. (ii) If the Rutherford’s atom model is true, the electron can revolve in orbits of all possible radii and hence it should emit continuous energy spectrum. However, the atoms like hydrogen possess line spectrum.

6. Bohr Model of the Hydrogen Atom Rutherford proposed the model of the atom that consists of the nucleus at the centre of an atom with electrons revolving around it. The nucleus-electron system is much like sun-planet system and is considered stable. But, there are some fundamental differences between the two system. While sun-planet system is governed by gravitational force, the nucleus-electron system is governed by coulomb’s law of force. We know that an object needs centripetal acceleration so that it can move in circular motion. Classical electromagnetic theory states that if a charged particle is accelerating, it emits radiation in the form of electromagnetic waves. According to this theory, electron, being a charged particle with continuously loose energy. This would make electron move spiraling into the nucleus. Also the frequency of the electromagnetic waves emitted by electrons in circular orbit is equal to the frequency of their respective revolutions. The radius of circular orbit of an electron is continuously decreasing as electron spiral inward. This continuously changes angular velocity and hence their frequency would change continuously because of which electrons would emit a continuous spectrum but in reality line spectrum is actually observed.

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Modern Physics-II So, classical physics is not sufficient to explain the atomic structure. It was Niels Bohr (1885-1962) who used the concept of quantized energy, and explained the model of a hydrogen atom in 1913. Bohr combined classical and early quantum concepts and proposed a theory in the form of three postulates. These postulates are: Postulate I: An electron in an atom could revolve in certain stable orbits without emitting radiant energy. Each atom has certain definite stable orbits. Electrons can exist in these orbits. Each possible orbit has definite total energy. These stable orbits are called the stationary states of the atom. Postulate II: An electron can revolve around the nucleus in an atom only in those stable orbits  whose angular momentum is the integral multiple of  (where h is Planck’s constant).  Therefore, angular momentum (L) of the orbiting electron is quantised.     where, n = 1, 2, 3 .....  Postulate III: An electron can make a transition from its stable orbit to another lower stable orbit. While doing so, a photon is emitted whose energy is equal to the energy difference between the initial and final states. Therefore, the energy of photon is given by, hv = Ei – Ef where Ei and Ef are the energies of the initial and final states. (Ei > Ef)

7. Mathematical Analysis of Bohr’s Theory for Hydrogen Atom. From the equations (i) & (ii) given below, we obtain various results.     and ... (i)           ... (ii)  (i) Velocity of electron in nth orbit: By putting the value of mvr in equation (i) from (ii), we get                  ⇒      .... (iii)    

 

  

where,

 ×  

    ×  ×   ×  × 

  ×     ≃  ×   where    ×   Speed of light in vacuum.

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Modern Physics-II (ii) Radius of the nth orbit : From equation (iii), putting the value of v in equation (ii), we get,   ×          ⇒       ... (iv)     ×  × ×    where,    × ×  × ×     ×   ≃ Å (iii) Total energy of electron in nth orbit: From eq. (i)              and              ∴ Total energy of the system = KE + PE  ⇒             









By putting the value of r from the equation (iv), we get          .... (v)     





 (iv) Time period of revolution of electron in nth orbit,     By putting the values of r and v, from (iii) and (iv),        ×     × ×   × ×     ×   where          × × × 

 

8. The Line Spectra of the Hydrogen Atom. The third postulate of Bohr’s model states that when an atom makes a transition from the higher energy state to the lower energy state, a photon is emitted with energy equal to the difference of energy of the state. Let the quantum number of higher energy state is ni and of lower energy state is nf and the frequency of emitted photon is ve then      The Ei and Ef can be defined using previous equation.

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Modern Physics-II

So,

              

or

             



                

  

  

 The above equation is the Rydberg formula. The term  is taken as Rydberg constant  R 



  ×  and  ≃Å        The value of Rydberg constant R is very close to the value  ×    obtained by Balmer using his empirical Balmer formula. The confirms the Bohr’s model of an atom. As we can see that nf and ni are integers, so the light which is radiated when an electron jumps between different energy levels, is of discrete frequencies moreover, an electron can comeback to its ground state or lower energy state. When an electron makes this transition from higher energy level to lower energy level, a photon is released. The energy of photon is equal to the energy difference in the energy levels.

In figure, n is principal quantum number and labels the stable states in the ascending order of energy. In the energy level diagram,   ∞ corresponds to the energy level of 0 eV. (Ionization energy state).

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Modern Physics-II

9. Hydrogen Spectrum (i) Lyman Series : In this series, the electron jumps from any outer orbit to first orbit. The general formula for wavelength of emitted radiation is given by            





       where    for first member of Lyman series    for second member of Lyman series .............. .......................... ............................ This series lies in ultraviolet region of the spectrum. Longest Wavelength of the Lyman Series : Longest wavelength (minimum energy) of this series is                           or, ∴        ×   Å   ×  × ×

  



Shortest Wavelength of Lyman Series (Series limit) : The shortest wavelength (maximum energy) of this series is obtained by putting   ∞           ∞       ×     Å ∴         × (ii) Balmer series : In this series, the electron jumps from any outer orbit to the second orbit, i.e.,          





Where





               for Ist member of Balmer series    for IInd member of Balmer series

   for IIIrd member of Balmer series ........ ............. .............. ............. .......... The first, second, third, ......... members of Balmer series are called       This series lies partly in the visible part of the spectrum. Longest wavelength of balmer series : Longest wavelength (minimum energy) of Balmer series corresponds to the transition                               ∴                      ×   Å. or,   ×  × ×

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Modern Physics-II Series limit of Balmer series or shortest wavelength (maximum energy) of Balmer series :      ∞ ∴





             ∞         ×     

(iii) Paschen Series : This series is emitted when the electron jumps from any outer orbit to the third orbit, i.e.,                    ∴       st where    for I member of Paschen series    for IInd member of Paschen series ........... ....... ....... ....... ........ ......... This series lies in the infra red region of the spectrum. Longest wavelength of Paschen series : For longest wavelength (minimum energy)                    Therefore                    or,        or,     ×   Å  × Series limit of Paschen series : For smallest wavelength (maximum energy) (or series limit),      ∞.               ∞         ×   Å 

10. Energy Levels: For H-atom, the energy of nth level is given by the relation : 

      ∴

             

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             

Modern Physics-II Taking these energies on a linear scale, horizontal lines are drawn which represent the energy levels of H-atom. The transition of electron shown on these energy levels gives the energy level diagram of H-atom. The arrows ending at n = 1, 2, 3, 4 and 5 represent the transitions responsible for Lyman, Balmer, Paschen, Brackett and Pfund series, respectively.

11. Excitation and Excitation Potential Excitation. The process of absorption of energy by an electron, so as to raise it from a lower energy level to some higher energy level, is called excitation. Excitation energy. The energy required, so as to raise an electron from its ground state to an excited state, is called the excitation energy. As obtained earlier, the energies of the electron in n = 1,2,3, ...... orbits of the hydrogen atom are E1 = -13.6 eV, E2 = -3.4 eV, E3 = -1.51 eV, ....... respectively. Therefore, energy required to raise the electron from n = 1 to n = 2 orbit, E2 – E1 = -3.4 - (-13.6) = 10.2 eV It is called first excitation energy of the hydrogen atom. Similarly, the second excitation energy of the hydrogen atom is given by E3 – E1 = -1.51 -(-13.6) = 12.09 eV Excitation potential. The potential difference, through which electron in an atom has to be accelerated so as to just raise it from its ground state to the excited state, is called excitation potential for that state. If an electron present in n = 1 state of the hydrogen atom is accelerated through a potential difference of 10.2 V, it will be raised to n = 2 state. Obviosuly, the first excitation potential of hydrogen atom is 10.2 V. Similarly, the second excitation potential of hydrogen atom is 12.09 V.

12. Ionisation and Ionisation potential Ionisation. The process of knocking an electron out of the atom is called ionisation. Ionisation energy. It is defined as the energy required to knock an electron completely out of the atom. When an electron is raised to the orbit   ∞ it will be completely out of the atom. Therefore, ionisation energy of hydrogen atom is equal to the energy required to raise it from orbit    to   ∞ i.e. ionisation energy  ∞          The ionisation energy is numerically equal to its ground slate energy. Since an accelerating potential of 13.6 V will raise the electron from orbit    to   ∞, the ionisation potential of hydrogen atom is 13.6 V.

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Modern Physics-II

13. Limitations of Bohr’s Atomic Model Following ae the limitations of bohr’s atomic model: (i) It explains the spectra of only hydrogen-like atoms i.e. atoms having only one electron, such as H, He+, Li++, etc. It fails to explain the spectra of multi-electron atoms. (ii) When a spectral line is observed under a powerful microscope, it is found to consist of a number of closely spaced lines. Bohr’s atomic model does not explain the fine structure of a spectral line. (iii) It does not explain the splitting of a spectral line into a number of spectral lines under the effect of magnetic field (Zeeman effect) and electric field (Stark effect). For example, when a sodium atom emitting radiation is placed in a magnetic field, each of the two yellow lines corresponding to wavelengths 5,890 Å further split into two fine lines. (iv) It does not tell about the relative intensities of spectral lines. (v) Bohr’s theory defines the orbits in which an electron can revolve very precisely i.e. defines both position and momentum with complete certainty. It is against the Heisenberg’s uncertainty principle. (vi) It considers an electron only as a particle, but electrons exhibit wave nature also. (vii) The researches show that the charge distribution due to electrons in different orbits is altogether different from the one assumed in Bohr’s model of atom

14. De-broglie’s Explanation of Bohr’s Second Postulate of Quantization de-Broglie explained second postulate of bohr’s atomic model by assuming an electron to a particle wave. Therefore, it should form standing waves under resonance condition. According to de-Broglie, for an electron moving in nth circular orbit of radius rn,          i.e., circumference of orbit should be integral multiple of de-Broglie wavelength of electron moving in nth orbit. As we know from previous chapter that de-Brogile wavelength,

    ⇒

    

13

or

    

Modern Physics-II

Which of the following is/are possible values of radius of stable orbit of hydrogen atom?   (a) (b)       (c) (d)          ∴ (a) & (c) are correct answers.

Example

Solution

15. Origin of Spectra The origin of spectra can be successfully explained on the basis of atomic theory. An atom consists of a central positive coe, called nucleus, which is surrounded by electrons revolving in various orbits according to a definite scheme. The energy associated with an electron in an outer is more that associated with an electron in an inner orbit. Therefore, when energy is given to an electron, it gets raised to some outer orbit. Since the number of electrons in an inner orbit must always be equal to 2 n2, the electron raised to the outer orbit returns back. When it does so, radiation having energy equal to the difference of energy associated with the two orbits is emitted.

16. Emission Spectra The light emitted by a source of light and resolved into its constituent wavelengths is called emission spectrum. Emission spectra are of the following three types : (i) Continuous emission spectrum. A continuous emission spectrum is one, which covers a wide range of wavelengths without any gap or discontinuity. The solar spectrum is an example of a continuous emission spectrum. In general, solids and liquids emit continuous spectra. (ii) Line emission spectrum. A line emission spectrum is one, which consists of bright lines (wavelengths) separated from each other by dark spaces. The line spectrum is emitted by the atoms of an element in gaseous or vapour form. (iii) Band emission spectrum. A band emission spectrum is one, which consists of a number of bright bands (group of wavelengths) of light, sharp at one end and fading off at the other end. The band spectrum is emitted by the molecules in gaseous form. These bands are sharply defined at one edge of the spectrum and shade off at the other edge.

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Modern Physics-II

17. Absorption Spectra When light from a source emitting full spectrum is passed through a substance (a solid or liquid in vapour state or a gas), the dark lines (or bands) appearing in the positions occupied by the bright lines (or bands) in the emission spectrum of the substance constitute the absorption spectrum of the substance.

18. X-rays and the Atomic Number Prof. Wilhelm Konrad Roentgen, a German scientist, in 1895, observed that when fast moving cathode rays strike a metal piece of high atomic weight and high melting point, a new kind of rays are produced. Since, nothing was known about these rays, Roentgen called these rays as X-rays. These are also called Roentgen rays. Roentgen was awarded noble prize in 1901 for this discovery. Quality Control in an X-ray tube : X-rays are of two types (i) Soft X-rays (ii) Hard X-rays. (i) Soft X-rays: X-rays having wavelength of 4 A or above (have longer wavelength), smaller frequency and hence smaller energy. These are called soft X-rays due to their low penetrating power. These are produced at comparatively low potential difference and high pressure. (ii) Hard X-rays: X-rays having low wavelength of the order of Å have high frequency and hence high energy. Their penetrating poewr is large. Therefore they are called hard X-rays. These are produced at comparatively low pressure and high potential difference. The wavelength of X-rays depend upon the kinetic energy of the electrons producing them and this kinetic energy depends upon the potential difference between the filament F and the target T (see below Coolidge tube figure). As the potential difference is increased, the wavelength of X-rays produced decreases, i.e., their penetrating power is increased or decreased by increasing or decreasing the potential difference applied between the ends of the tube.

15

Modern Physics-II Properties of X-rays Properties of X-rays may be divided into two headings : (i) Properties similar to light rays: (a) These are invisible. (b) These travel with the speed of light. (c) These travel in straight line. (d) These undergo reflection, refraction, interference, diffraction and polarisation. (e) The wavelength of X-rays is very small compared to the wavelength of light. Hence, these carry much more energy. (This is the only difference between X-rays and light). The wavelength of light rays range from 4000 to 7500 Å, whereas that of X-rays is from 100 Å to 1 Å. (f) These produce illumination after falling on fluorescent materials. (g) X-rays are not deflected by electric and magnetic fields. (h) These show continuous spectrum. Hence we conclude that like light rays, X-rays are electromagnetic waves. These show all important properties of light rays. (ii) Properties of X-rays similar to cathode rays: (a) X-rays ionise the gases through which these pass. (b) These penetrate through different depths into different substances, e.g., wood, cardboard, thin metal sheets. (c) If objects are placed in their path, these cast their shadows. (d) These affect photographic plates. (e) These are very active and may eject electrons from metals e.g., these show photoelectric effect. (f) Long exposure of both cathode rays and X-rays is injurious to human body.

Absorption of X-rays If a beam of X-rays of intensity  passes through a length dx of any material, its intensity is decreased by dl. For any given material, the amount of absorbed intensity is  dx, where  is called the absorption coefficient. Then,           or on integration we get, log     ×   ... (1) where ‘c’ is constant of integration When ×   then    hence c = log l0 Eq. (1) becomes log     ×  log l0 log   log     × or log (lll0)    ×     or where,  is the intensity of incident X-rays and,  that of emergent X-rays.

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Modern Physics-II

19. Spectrum of X-rays The X-rays produced in Coolidge tube gives a continuous spectrum crossed over by characteristic lines. It means that the X-rays contain a wide range of wavelengths whose intensities are so intermixed that these cannot be distinguished from the another. The variation of intensity with wavelength in the spectrum of X-rays is shown in figure

The resulting curve shows that there is a lowest wavelength limit min for X-rays. The wavelengths shorter than min do not exist in the X-rays emitted from the Coolidge tube. X-ray spectrum is of two types: (a) Continuous X-ray spectrum (b) Characteristic X-ray spectrum (a) Continuous X-rays : The emission of X-rays by electrons striking the target and the lowest wavelength limit can be explained by the quantum theory. According to this theory, emission of X-rays takes place in the form of small bundles of energy, called ‘photons’. The energy of each photon is hv, where v is the frequency of X-radiation and h is Planck’s constant. When a voltage V is applied across the X-ray tube, the electron emitted by the filament will reach the target with energy eV. When the electron strikes an atom of the target, it loses a part or whole of its energy. This lost energy is obtained in the form of an X-ray photon, given by the relation:         The electron rarely loses whole of its energy in a single collision. Generally it undergoes a sequence of collisions with atoms of the target before coming to rest, thus emitting photons of smaller and smaller energies i.e., of longer and longer wavelengths. This accounts for the production of a continuous range of X-rays wavelengths, above a definite limit. Maximum frequency of continuous X-rays : When the electron loses the whole of its energy in a single collision with the target atom, an X-ray photon of maximum energy hvmax is emitted. Thus, for an accelerating voltage V, the maximum X-rays photon energy is given by hvmax = eV. Consequently, the maximum frequency of X-rays produced by electron of energy eV vmax   

17

Modern Physics-II Minimum wavelength of continuous X-rays: The minimum wavelength corresponding to maximum frequency vmax is given by     min        Å  ∵max   max   





where c is the speed of light. Thus, the minimum wavelength limit min  is inversely proportional to the accelerating potential (V). (b) Characteristic X-rays : The spectrum of X-rays is a continuous spectrum crossed over by distinct spectral lines whose frequencies are the characteristic of the material of the target and the line spectrum is called characteristic X-ray spectrum of the material of the target. The radiation forming the line spectrum is called characteristic X-rays. Characteristic X-ray spectrum is simple one. Most element show only K-series and L-series in their characteristic X-ray spectra. Wavelengths of K-series are generally less than Å and those of L-series are nearly 10 times longer. First, second, third,...members of the K and L series are      and     respectively. Elements having atomic number greater than 66 show further series known as M and N series. Origin of characteristic X-rays An atom consists of a central positive nuclear us and electrons revolving around it in orbits of definite radii. These shells are called K, L, M, N,..... shells, K shell being the innermost shell. The electrons in the K-shell are most tightly bound to the nucleus, and maximum force is requird to elect them. Less energy is required to eject the electrons from the L-shell and still smaller energy for M-shell and so on.

In an X-ray tube an electron strikes the target with such a large velocity that it can penetrate well into the atoms of the target and eject a K-electron. A vacancy is created in the K-shell, and it is immediately filled by an electron from L-shell. The balance of energy (which is equal to the difference in energies of the two shells) is emitted out as an X-ray photon. This line corresponds to  line in K-series of X-ray spectra.

18

Modern Physics-II Similarly, if the vacancy of K-shell is filled by the jumping of electrons from the M, N... shells,    lines of this series are emitted. If the electron striking the target ejects an electron from the L-shell, of the target atom, an electron from the M, N ... shell, jumps to occupy this shell. Thus X-ray photons of less energy are emitted. These photons form L-series. In a similar way the formation of M, N... series can be explained.

20. Moseley’s law Moseley (1913) made an extensive study of the characteristic X-ray spectra of a number of heavy elements, and observed a simple relationship between them. He found that the spectra of different elements are very similar, and with increasing atomic number Z, the spectral lines merely shift towards shorter wavelength or higher frequencies. He plotted a graph of the K-series between the square-root of frequency (√f) and atomic number (Z) and found it very approximately to be a straight line. He, therefore, concluded that the square-root of the frequency of a K-line is closely proportional to the atomic number of the element.  ∝   or        This is called Moseley’s law and may be expressed as  where Z is the atomic number of the element, and K and b are constants for a given transition of the K-series. The law is applicable to other series also but with different values of K and b.  ×  and    For  X-ray, take    

Hence, the frequency of a  X-ray is    ×    

19

Modern Physics-II  and the atomic weight, there is an appreciable If, however, a graph be plotted between  departure from straight line. Moseley concluded that it is the atomic number, and not the atomic weight, which is more fundamental with regard to the emission of characteristic X-ray. Moseley pointed out that the elements in the periodic table must be arranged in the order of increasing atomic number instead of atomic weight. From this point of view he changed the position of certain elements in the Mendeleev’s Periodic Table. When these changes were made, the anomalies of the Mendeleev’s table disappeared. While arranging the elements in the increasing order of atomic number, Moseley had to leave certain gaps such as at Z = 43, and 72. He pointed out that these elements may be discovered later on and were subsequently discovered. These elements are technetium and hafnium, respectively.

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Modern Physics-II

AIIMS Exercise (1)  1. In Thomson’s experiment of finding  for electrons, beam of electrons is replaced by that of  muons (particle with same charge as that of electrons but of mass 208 times that of electron). No deflection condition in this case is satisfied if : (a) B is increased 208 times (b) E is increased 208 times (c) B is increased 14.4 times (d) None of these

2. The distance between the deflecting plates in Thomson’s experiment is 2 cm and the deflecting  and potential is 3000 V. A magnetic field of 2.5 × 10‒3 Wb/m2 is applied perpendicular to    . The accelerating potential is 10 kV. Then  of the electrons is :  (a) 3.6 × 1011 Ckg‒1 (b) 4.2 × 1011 Ckg‒1 (c) 1.8 × 1011 Ckg‒1 (d) 1.6 × 1019 Ckg‒1 3. An electron, accelerated though a potential difference of 900 V, acquires a velocity of 1.8 × 107 m/s. then the specific charge of he electron is : (a) (b) (c) (d)

1.8 2.8 3.8 0.8

× × × ×

1011 1011 1011 1011

Ckg‒1 Ckg‒1 Ckg‒1 Ckg‒1

4. A beam of electrons each of mass m and charge e and moving with a velocity of 3 × 107 ms‒7 is deflected by 1.8 mm in travelling a distance of 10 cm through an electric field of 1800 V/m.  If the electric field is perpendicular to their initial velocity, then  of the electrons is :  (a) 3 × 1011 Ckg‒1 (b) 1.8 × 1011 Ckg‒1 (c) 0.9 × 1011 Ckg‒1 (d) 2 × 1011 Ckg‒1

21

Modern Physics-II 5. A beam of electrons travelling with a velocity of 2.5 × 107 ms‒1 passes through crossed electric  = 0.04 tesla, calculate the magnitude of and magnetic fields perpendicular to each other. If    so that the electron beam remains undeflected : (a) 104 V/m (b) 106 V/m

(c) 105 V/m (d) 103 V/m

6. In J.J. Thomson’s parabolic method for positive ray analysis, two identical parabolas are obtained for the same electric field. If the magnetic fields are respectively 6T and 4T, compare the masses of the particles : (a) 1 : 4 (b) 4 : 1

(c) 9 : 4 (d) 8 : 9

7. The radius of a nucleus with atomic number 7 is 2 fermi. The radius of the nucleus with atomic number 189 is : (a) 3 fermi (b) 4 fermi

(c) 5 fermi (d) 6 fermi

8. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 : 1. The ratio of their nuclear sizes will be :   

(a)    (b)   

  

  

(c)     

(d)    

9. The binding energy per nucleon of deutron 1H2 and helium nucleus (2He4) is 1.1 MeV and 7 MeV respectively. If two deutrons nuclei react to form a single helium nucleus, then the energy released is : (a) 13.9 MeV (b) 26.9 MeV

(c) 23.6 MeV (d) 19.2 MeV

10. In hydrogen atom spectrum, frequency of 2.7 × 1015 Hz of electromagnetic wave is emitted when transmission takes place from 2 to 1. If it moves from 3 to 1, the frequency emitted will be : (a) 3.2 × 1015 Hz (b) 32 × 1015 Hz (c) 1.6 × 1015 Hz (d) 16 × 1015 Hz

22

Modern Physics-II 11. The ionisation potential of Li++ ion is 122.4 volt. The first excitation potential is : (a) 122.4 V (b) 91.8 V

(c) 61.2 V (d) 30.6 V

12. The hydrogen atom makes a transition from n = 4 to n = 1, then the recoil momentum of  H-atom in units of  is given by : 

 

(a) 13.6 (b) 12.75

(c) 0.85 (d) 22.1

In each of the following questions, a statement of Assertion (A) is followed by a corresponding statement of Reason (R). Of the following statements, choose the correct one. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. (e) Both A and R are false. 13. (A) : X-rays can ionise a gas. (R) : Their wavelength is very small as compared to light waves. 14. (A) : In Millikan’s oil drop experiment for the determination of charge on an electron, the oil drops may be of any size. (R) : In this experiment, the charge on the oil drops is measured by determining their terminal velocities.

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Modern Physics-II

01. Introduction We shall now study the constituents of the nucleus and how they are held together. We shall discuss various properties of nuclei such as size, mass, density and stability of nuclei and associated nuclear phenomena such as radioactivity, nuclear fission and nuclear fusion.

02. Atomic Masses The unit in which atomic and nuclear masses are measured is called atomic mass unit (a.m.u.) One atomic mass unit is defined as 1/12/ th of the mass of an atom of 6C12 isotope. As Avogadro’s number   ×  ∴ Mass of  ×  atoms of       Mass of one atom of      ×   By definition, 1 a.m.u.   × mass of one atom of         ×    ∴ 1 a.m.u. =  ×    ×    ×    As an atom of 6C12 contains 12 nucleons, therefore, one a.m.u. represents the average mass of a nucleon and is denoted by u. In terms of this unit, mass of an electron (me) = 0.00055 u mass of proton (mp) = 1.0073 u mass of a hydrogen atom = me + mp = 1.0078 u mass of a neutron (mn) = 1.0086 u. Electron volt (eV). It is the unit of energy. One electron volt is the energy gained by an electron, when accelerated through a potential difference of one volt. Relation between a.m.u. and MeV According to Einstein, mass energy equivalence is represented by E = mc2  Taking m = 1. a.m.u. =  ×    and    ×  we get,    ×     ×     ×    ×        ∴   ×  Hence, 1 a.m.u. ≃  , which is used as a standard conversion.

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Modern Physics-II

03. Neutron A neutron is a neutral particle carrying no charge and having mass roughly equal to the mass of a proton. Now the mass of a neutron is known to a high degree of accuracy and is equal to mn = 1.00866 u = 1.6749 × 10-27 kg

04. Composition of Nucleus Nuclei are composed of protons and neutrons. Following points should be remembered about protons and neutrons. Proton :

(i) (ii) (iii) (iv)

The nucleus of hydrogen atom is called proton. It is an essential constituent of all nuclei. It carries positive charge =  ×   Its mass   ×        times the mass of an electron (v) Due to its positive charge, protons are not used as projectile in nuclear fission. (vi) It is stable even outside the nucleus. (vii) Its qua가 content is uud. Neutron:

(i) (ii)

Neutron was discovered in 1932 by James Chadwick. It has no charge, hence it is a neutral particle.

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Modern Physics-II

(iii) Its mass =  ×    ≃ nearly equal to that of proton =  times the mass of an electron =   (iv) It is an essential constituent of all nuclei except hydrogen (1H1). (v) It has very high penetrating power because it can neigher be attracted nor repelled by the nucleus. (vi) A fast moving neutron can be slowed down by materials called moderator. Examples of moderators are heavy water, graphite, paraffin wax etc. Neutrons are in thermal equilibrium with the molecules of the moderator when both attain the same energy. The kinetic energy of a thermal neutron is about 0.04 eV. (vii) Thermal neutrons are used as projectiles in a nuclear fission. (viii) Neutron has very low ionising power. (ix) A free neutron (i.e., outside the nucleus) is unstable and decays into a proton, an electron and an antineutrino.        (antineutrino)  → 

(x) (xi)

(proton) (electron) The half life of a free neutron is about 12 minute. The mean life is about 1000 s. Neutron inside the nucleus is stable. The quark content of a neutron is udd.

Nucleons : Protons and neutrons taken together are called nucleons. We regard, a proton and a neutron as two different charge states of the same particle, called “nucleon”. Mass Number (A) : The total number of nucleons (=neutrons + protons) in the nucleus of the atom is called mass number of the atom. It is denoted by A. It is different from the mass of the atom which includes the total mass of neutrons plus protons plus electrons. Atomic Number (Z) : The total number of protons in the nucleus of an atom is called atomic number of that atom. It is denoted by Z. The arrangement of various elements in the periodic table is according to their atomic numbers. Let N be the number of neutrons in the nucleus of an atom with A as the mass number and Z as the atomic number. Then A = N + Z or N = A – Z ...(i) The nucleus of an atom is denoted by the chemical symbol (X) of the atom with the atomic number Z as the subscript and mass number A as the superscript. hence we write or represent a nuclide/nucleus as 

 Symbol)A or  For example, nucleus for uranium is 92U238 It contains 92 protons and 238-92=146 neutrons. Z(Chemical

Nuclear charge : Total charge of the nucleus is equal to the total charge on all the Z protons in it = Ze where    ×  

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Modern Physics-II

05. Nuclear Size Experimental measurements show that volume of a nucleus is proportional to its mass number  A. If R is the radius of the nucleus assumed to be spherical*, then its volume     ∝     ∝    or or ...(i)





where  is an empirical constant whose value is found to be  ×    As a is different for different elements, therefore, atomic nuclei of different elements have different sizes.

06. Nuclear Density Density of nuclear matter is the ration of mass of nucleus and its volume. If m is average mass of a nucleon and R is the nuclear radius, then, mass of nucleus = mA, where A is the mass number of the element.     Volume of nucleus                    As density of nuclear matter   ∴             Thus,     As m and R0 are constants, therefore density  of nuclear matter is the same for all elements.    ×   Using    ×     × ×    ×  we get,        × ×  which is very large as compared to density of ordinary matter. For example, for water,  ≈   and for air,     Hence matter in the nucleus is very densely packed. NOTE ☞

Note that sizes of nuclei are different, the heavier nuclei being bigger than the lighter ones. But density of nuclear matter is same for all nuclei. Value of nuclear density is far more greater that the atomic density.

Example

What is the ratio of nuclear densities of the two nuclei having mass numbers in the ratio 1:4? Ratio of nuclear densities is 1:1 because nuclear density is independent of mass number.

Solution

27

Modern Physics-II What is the nuclear radius of

Example Solution

  

125

Fe if that of

27

AI is 3.6 fermi?



      



       





        

07. Isotopes Nuclei having the same nuclear charge (Ze) but different mass numbers (A) are called isotopes. Examples: (1)            are isotopes of carbon. (2)      are isotopes of oxygen. (3)  and  are isotopes of copper. (4) Hydrogen has three isotopes given below: (a) Hydrogen (1H1) : It has the simplest nucleus with only one nucleon called proton. It has no neutron. Its relative abundance is 99.85%. (b) Deuterium (1H2) : Its nucleus is called deuteron with one proton and one neutron. (c) Tritium (1H3) : It nucleus is called triton with one proton and two neutrons. It is unstable and hence not found in the nature. (5) The element gold has 32 isotopes from A=173 to 204. Isobars Nuclei having the same mass number A but different nuclear charge (Ze) are called isobars. Examples : (1) 11Na22 and 10Ne22 are isobars with the same mass number 22. (2) 11Na24 and 12Mg24 are isobars with the same mass number 24. Isobars are atoms of different elements and hence they have different chemical properties. Isotones Nuclides having the same number of neutrons are called isotones. For example 17Cl37 and 19K39 are with 20 neutrons. Nuclide : A single nuclear species with specific values of both Z and N called a nuclide. Example

The three stable isotopes of neon Ne20, Ne21 and Ne22 have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 respectively. Obtain the average atomic mass of neon.

Solution

Average atomic mass of neon.

     ≠   

      ×   ×   ×            

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Modern Physics-II

08. Mass-Energy Relation Einstein was the first to establish the equivalence of mass and energy through the famous relation       ×   speed of light in vacuum where The relation means that when a certain mass m disappears, an equivalent amount of energy E appears and vice-versa. In nuclear and elementary particle interactions, both the conversion of mass into energy and the conversion of energy into mass take place. Therefore, the two classical laws of conservation of mass and conservation of energy have been unified into one law-the law of conservation of mass energy. According to this law, the sum of the mass-energy of a system of particles is the same, before and after an interaction. Express 1 mg mass equivalent in eV.

Example Solution



      × joule   × ×     ×     ×

09. Mass Defect ∆ Binding Energy (Eb) and Binding energy per nucleon (Ebn) The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the constituents into free nucleons. The energy with which nucleons are bounded together in a nucleus is called as Binding Energy (B.E.) or Eb. In order to free nucleons from a bounded nucleus, this much of energy (=B.E.) is to be supplied. It is observed that the mass of a nucleus is always less than the mass of constituent (free) nucleons. This difference in mass is called as mass defect and is denoted as ∆m. If mn : mass of a neutron ; mp : mass of a proton m(Z,A) : mass of bounded nucleus; M(Z,A) : mass of the atom             ∆         Then,                  This mass-defect is in form of energy and is responsible for binding the nucleons together. From Einstein’s law of inter conversion of mass into energy:    (c: speed of light; m : mass) ⇒ Binding energy  ∆ Generally, ∆ is measured in amu units. So let us calculate the energy equivalent to 1 amu. It is calculated in eV (electron volts;    ×    

 × ×  ×  ×  ×    ×∆ ∈

 ≡      ×       ⇒

There is another quantity which is very useful in predicting the stability of a nucleus called



as Binding energy per nucleon  and is given by :    

29

Modern Physics-II Example

The nucleus of

  

has 8 protons and 8 neutrons. Calculate binding energy and binding

energy per nucleon fopr oxygen atom. Given :

          Actual mass of    atom=15.99493  Expected mass of nucleus       ×   ×  

Solution

Actual mass of

  

  

∆m = (Expected mass of nucleus)-(Actual mass of

  

atom-mass of   )

∆          ×∆   ×           and 

Binding Energy Curve

Observations: F  increases on an average and reaches a maximum of about  for A=56. F for more heavy nuclei  decreases slowly as A increases. For the heaviest natural element  it drops to about 7.6MeV. F Nuclei in the region of mass number 30 Gravitational force (c) Nuclear forces are independent of charge. The nuclear force between two protons is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces. (d) Nuclear forces are always attractive in nature. (e) Due to short ranged nature of nuclear force, another important property known as “saturation property of nuclear force” comes into picture. In the range, 30 1, the reaction rate increases exponentially to explode. For K < 1, the reaction rate reduces rapidly. (iv) Coolant : The energy released inside the reactor in the form of heat is removed by coolant. For this purpose air, ice cold water, molten sodium or CO2 is circulated around the reactor core area which withdraws the heat produced in the core. This heat is utilised for producing steam which is then used to drive turbines for generating electricity.

41

Modern Physics-II (v)

Shielding Wall : Various types of harmful intense rays are emitted from the reactor. To protect the human beings from these rays, the reactor is surrounded by 7 to 8 ft thick concrete walls. Unlike the waste of thermal power stations the waste of a nuclear power is highly radioactive and extremely hazardous to all forms of life. Hence, elaborate safety measures are required.

Breeder Reactors The reactors, which can produce fuel more than they use, are called “breeder reactors”. We have known that not only U236 but Pu240 is also highly fissionable. Pu240 is obtained by bombarding Pu239 with thermal neutrons. But Pu239 is not a naturally occurring isotope. However, U238 can capture a neutron to produce Pu239 which can be used as fuel. 238 1 92U + 0n

239 92U

β-

239 93Np

β-

94Pu

239

If more than one neutron can be absorbed by U238 rods per fission, then we produce more fuel in the form of Pu239 as we consume in the form of U235. Thus, apart from energy, these reactors give us fresh nuclear fuel which often exceeds the nuclear fuel used.

42

Modern Physics-II

AIIMS Exercise (1) 1. The total energy of an electron in hydrogen atom in the second quantum state is ‒E2. the total energy of the electron of He+ ion in the 3rd quantum state is :  (a)      (b)    

 (c)      (d)    

2. The hydrogen atom in its ground state is excited by means of radiations of wavelengths 975Å. The lines possible in spectrum are : (a) 6 (b) 5

(c) 7 (d) 4

3. The energy of excited hydrogen atom is ‒3.4 eV. The angular momentum of the electron according to Bohr theory (in J-s) is : (a) 1.1 × 10‒32 (b) 1.12 × 10‒34 (c) 2.11 × 10‒34 (d) 2.1 × 10‒36 4. If binding energy of deuterium is 2.23 MeV, the mass defect in amu is : (a) ‒0.0024 (b) ‒0.0012 (c) 0.0012 (d) 0.0024 5. The shortest wavelength of Lyman series is equal to the shortest wavelength of Balmer series of a hydrogen like atom of atomic number Z. Then Z is : (a) 4 (b) 2

(c) 3 (d) 6

43

Modern Physics-II 6. An electron in H-atom makes a transition from n = n1 to n = n2. The time period of the electron in the intial state is eight times that in the final state. The possible values of n1 and n2 are : (a) n1 = 6, n2 = 2 (b) n1 = 8, n2 = 2 (c) n1 = 8, n2 = 1 (d) n1 = 6, n2 = 3 7. The angular momentum of an electron in a hydrogen atom is proportional to :  (c)   (d) 

 (a)     (b)  

8. Magnetic field at the centre (at nucleus) of hydrogen like atom (atomic number Z) due to motion of electron in the nth orbit is proportional to :  (a) 

(c)  



 (b)  

(d)  





9. The number of α-particles emitted per second by N-atoms of a radioactive element is n. Then the half-life period of the element is :  (a)  sec 



(b)  sec 



(c) 0.693  sec   (d) 0.693  sec 

10. Two radioactive materials x1 and x2 contain same number of nuclei. If 6λ s‒1 and 4λ s‒1 are the decay constants of x1 and x2 respectively, the ratio of number of nuclei undecayed of x1 to that  x2 will be  after a time? 

 

 (a)  sec   (b)  sec 

44

 (c)  sec   (d)  sec 

Modern Physics-II 11. The element with atomic number 11 emits Kα X-ray of wavelength λ, then the atomic number of the element which emits Kα X-rays of wavelength 4λ is : (a) 11 (b) 44

(c) 6 (d) 5

12. The wavelength of Kα-line from an element of atomic number 60 is λ. Then the wavelength of Kα-line of an element of atomic number 30 is : (a)   (b)  

(c)   (d)  

In each of the following questions, a statement of Assertion (A) is followed by a corresponding statement of Reason (R). Of the following statements, choose the correct one. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. (e) Both A and R are false. 13. (A) : Neutrons penetrate matter more readily as compared to protons. (R) : Neutrons are slightly more massive than protons. 14. (A) : Positive value of packing fraction implies a large value of binding energy. (R) : Packing fraction is the difference between the mass of the nucleus and the mass number of the nucleus.

45