Chapter

If we provide energy to the electron of atom then there \(i\) a possibility that is it is excited to higher energy level.
This process can be done in two ways :

• By Absorption of photons

• By collision with other atoms and electrons

By Absorption of photons

If an electron is to absorb a photon the energy \(hv\) of photon must be equal to the energy difference \(\Delta E\) between the initial energy level of the electron and a higher level.

It means if we consider the case of H -atom then this atom can absorb only certain specific energy photons which are \(10.2ev,12.75ev\) etc.

The electron of H -atom can not absorb photon of energy 11 ev but this electron can absorb any photon of energy greater then 13.6 eV or more specifically ionization energy of atom. After absorbing energy more than 13.6 eV , rest of the energy may appear as kinetic energy of the free electron.

Absorption spectrum :

When an atom absorb a photon whose energy is exactly equal to the difference of ground state and any of the excited states, this wavelength corresponds line of absorption spectrum.

Note :

• An atom will absorb energy from its ground state only.

• Wavelengths of absorption spectrum can be determined as

\[\frac{1}{\lambda} = {RZ}^{2}\left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right)\ n = 2,3,4,\ldots\ldots\ldots\]

• Number of lines in abortion spectrum between \(n = 1\) and \(n = n\) level will be ( \(n = 1\) )

By collision with other atoms and electrons

The electron of an atom may also absorb energy during collisions, and may be excited to a higher energy state. During collisions of atoms and electrons the loss of energy must be used to excite the atoms as at atomic level then is no significance of Thermal energy and rise of temperature. We can not estimate that what type of collision must occur but we can always analyze the possibilities during a collision.
(i) If loss of KE during collision is not sufficient to excite the atom then collision must be perfectly elastic.
(ii) The collision may be inelastic or perfectly inelastic only if loss of KE is exactly equal to any of the excitation energy of the atom.
During a collision maximum loss of KE can be calculated using center of frame. i.e. KE with respect to COM can be lost.

\[{KE}_{\text{system, COM}\text{~}} = \frac{1}{2}\mu{\text{ }V}_{\text{rel}\text{~}}\]

Where \(\mu \rightarrow\) Reduced Mass of system
\(V_{rel} \rightarrow\) relative velocity of the atoms
Consider an H -atom at rest and a neutron with \(KE = 25eV\) is going to collide with the H -atom. Mass of neutron and H -atom can consider as same.

\[\begin{matrix} & {KE}_{\max\text{~}\text{lost}\text{~}} = \frac{1}{2}\mu v_{\text{rel}\text{~}}^{2} \\ & \ = \frac{1}{2}\left\lbrack \frac{\text{ }m \cdot \text{ }m}{\text{ }m + m} \right\rbrack v_{0}^{2} \\ & \ = \frac{1}{2}\lbrack KE\text{~}\text{of neutron}\text{~}\rbrack \\ & \ = 12.5ev \end{matrix}\]

\(1^{\text{st}\text{~}}\) Possibility: Pertectly elastic collision and no excitation

• This is a possible case in every collision if no loss takes place, no excitation to will be there.
  \(2^{\text{nd}\text{~}}\) Possibility : Inelastic collision, H -atom excited to \(n = 2\) for this \(\Delta E = 10.2ev\) is required and it is less than \({KE}_{\text{max loss}\text{~}}\) hence a possible case.
  \(3\ ^{\text{rd}\text{~}}\) Possibility: In elastic collision, H - atom excited to \(n = 3\) for this \(\Delta E = 12.09ev{KE}_{\text{max lass}\text{~}}\) also passible
  \(4\ ^{\text{th}\text{~}}\) Possibility: In elestic collision, H -atom excited to \(n = 4\) for this \(\Delta E = {KE}_{\text{max loss}\text{~}}\ldots.\). not Possible
  \(5\ ^{\text{th}\text{~}}\) Possibility: Pertectly in eleastic collision.
  In this case \(\Delta E = {KE}_{\text{max loss}\text{~}} = 12.5ev\ldots\). not Possible as 12.5 ev is not an excitation energy.
  Note : In all the possibilites, apart from loss rest of the KE will shared by H -atom and newtron. Which can be calculated using momentum - consevation case.

Practice Exercise

Q. 1 State whether following statements are true/False :
(a) A neutron moving with \(KE = 20ev\) collides with an H -atom at rest. The collision must be perfectly elastic.
(b)A neutron moving with \(KE = 30ev\) collides with an H -atom at rest. The collision may be perfectly in elastic.
(c) An H -atom moving with \(KE = 25.5ev\) collides with another H -atom at rest. The collision may be perfectly inelastic.
Q. 2 A neutron having kinetic energy 12.5 eV collides with a hydrogen atom at rest. Neglect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.
Q. 3 A neutron moving with a speed v strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron \(=\) mass of hydrogen \(= 1.67 \times^{- 27}\text{ }kg\).

Answers

Q. 1 (a) T (b) \(T\ \) (c) \(T\ \) Q. \(2\ \) zero \(\ \) Q. \(3\ 3.13 \times 10^{4}\text{ }m/s\)

Electrons excited to higher energy stay their only for \(10^{- 8}\text{ }s\) then they make transition to any lower state by emitting photons and finally come to a ground state. Energy of photons emitted is equal to the difference of energy of the levels. While coming down they may emit photons of various wavelengths which corresponds to several spectral series.

Emission spectrum :

When an electron in excited state makes a transition to a lower state, a photon is emitted. Collection of these photon wavelengths in called emission spectrum.

Hydrogen atom (or hydrogen like hydrogen atoms) consists of only one electron but we get a number of spectral lines in its spectrum (emission).

1. Lyman series: The spectral lines of this series correspond to the transition of an electron from some higher energy state to the inner most orbit ( \(n = 1\) i.e. ground state).
   For Lyman series, \(n_{1} = 1,n_{2} = 2,3,4,\ldots\ldots\)
   So, \(\ \frac{1}{\lambda_{\text{Lyman}\text{~}}} = R\ Z^{2}\left( \frac{1}{l^{2}} - \frac{1}{n_{2}^{2}} \right)\)
   for hydrogen atom, \(Z = 1\)

\[\frac{1}{\lambda_{Lyman}} = R\left( \frac{1}{l^{2}} - \frac{1}{n_{2}^{2}} \right)\]

2. Balmer series : The spectral lines of this series correspeond to the trasition of an electron from higher energy state to an oribit having \(n = 2\),
   For Balmer serise \(n_{1} = 2,n_{2} = 3,4,5,\ldots\ldots.\).
   The wave numbers and the wavelengths of spectral lines constituting the Balmer series are given by,

\[\frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{n_{2}^{2}} \right)\]

3. Paschen series: The spectral lines of this series correspand to the transition of an electron from some higher energy state to an orbit having \(n = 3\).
   For Paschen series, \(n_{1} = 3,n_{2} = 4,5,6\),
   The wave numbers and the wavelengths of spectral lines constituting the paschen series are given by,

\[\frac{1}{\lambda} = R\left( \frac{1}{3^{2}} - \frac{1}{n_{2}^{2}} \right)\]

Paschen series is so named because it was discovered by paschen. Just like other series, this eries was first predicted by Bohr.
4. Barcket series: The spectral line of this series correspond to the transition of an electron from a higher energy state to the orbit having \(n = 4\).
For this serie, \(n_{1} = 4,n_{2} = 5,6,7,\ldots\ldots\)

\[\frac{1}{\lambda} = R_{\infty}\left( \frac{1}{6^{2}} - \frac{1}{n_{2}^{2}} \right)\]

5. Pfund series : The spectral line of this series correspond to the transition of an electron from a higher energy state to the orbit having \(n = 5\).
   For the series, \(n_{1} = 5\) and \(n_{2} = 6,7,8,\ldots\ldots.\).
   The wave number and the wavelength of the spectral lines constituting the Pfund series are given by.

\[\frac{1}{\lambda} = R_{\infty}\left( \frac{1}{5^{2}} - \frac{1}{n_{2}^{2}} \right)\]

6. Humphery series : For this series, \(n_{1} = 6\) and \(n_{2} = 7,8,9\),

For this series, \(n_{1} = 6\) and \(n_{2} = 7,8,9\), \(\_\_\_\_\)
The wave number and the wave lengths of the spectral lines constiting the Humphery series are given by,

Spectral lines originate in transitions between energy levels.
Shown are the spectral series of hydrogen. When \(n = \infty\), the electron is free.
Note :

• While coming down to ground state a single atom in \(n = n\) state may emit a maximum ( \(n - 1\) ) photons.

• First line of a series corresponds to lowest energy photon emitted e.g. first line of blamer series corresponds to transition from \(n = 3\) to \(n = 2\).

• Series limit corresponds to maximum energy photon emitted e.g. series limit of blamer series corresponds to transition from \(n = \infty\) to \(n = 2\).

• Maximum number of lines in emission spectrum of a gas which is excited to a level \(n = n\) will be \(\ ^{n}C_{2}\) i.e. \(n(n - 1)/2\).

• For an atom in quantum state \(n = n\)

Max energy photon will be emitted for a transition from \(n = n\) to \(n = 1\).
Min energy photon will be emitted for a transition from \(n = n\) to \(n = n - 1\).

Mass of the nucleus is comparable to mass of electron

If mass of the nucleus is comparable to mass of electron, then the electron and nucleus revolve in coplanar concentric circular paths of radii \(r_{1}\) and \(r_{2}\) about their common centre of mass. The electrostatic force of attraction provides them necessary centripetal force. They revolve with the same angular velocity and their sense of rotation is also same.
from the figure,

\[\begin{array}{r} \begin{matrix} & r_{1} + r_{2} = r_{n} \\ \text{~}\text{and}\text{~} & {Mvr}_{1} = {mv}_{2} \end{matrix}\#(i) \end{array}\]

or \(\ \frac{r_{1}}{r_{2}} = \frac{m}{M}\)
from equation (i) and (ii),

\[r_{1} = \left( \frac{m}{m + M} \right)r_{n};\ r_{2} = \left( \frac{m}{m + M} \right)r_{n}\]

Let their angular velocities be \(\omega_{1}\) and \(\omega_{2}\) and electrostatic force of attraction between them be F , then,

\[\begin{array}{r} F = M\omega_{1}^{2}r_{1}\#(forM) \end{array}\]

and

\[\begin{array}{r} F = m\omega_{2}^{2}r_{2}\#(form) \end{array}\]

So, \(\ M\omega_{1}^{2}r_{1} = m\omega_{2}^{2}r_{2}\) but \({Mr}_{1} = {mr}_{2}\) [from equation (ii)]
So, \(\ \omega_{1}^{2} = \omega_{2}^{2}\) or \(\omega_{1} = \omega_{2} = \omega\) (say)
So they revolve with same angular velocity about their common centre of mass.

Now let us see why ' \(m\) ' is replaced by the reduced mss \(\mu\) when motion of nucleus is also to be considered. Centripetal force to the electron is provided by the electrastic force, so,

\[{mr}_{2}\omega^{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{{Ze}^{2}}{r_{n}^{2}}\ \text{~}\text{or}\text{~}\left( \frac{Mm}{M + m} \right)r_{n}^{3}\omega^{2} = \frac{{Ze}^{2}}{4\pi\varepsilon_{0}}\]

or

\[\begin{array}{r} \mu r_{n}^{3}\omega^{2} = \frac{{Ze}^{2}}{4\pi\varepsilon_{0}}\#(iii) \end{array}\]

where,

\[\mu = \frac{1}{m} + \frac{1}{M} = \frac{Mm}{M + m}\]

Now, moment of inertia about the common centre of mass,

\[I = Mr_{1}^{2} + mr_{2}^{2} = M\left( \frac{m}{m + M} \right)^{2}Mr_{1}^{2} + mr_{2}^{2} + m\left( \frac{M}{m + M} \right)^{2}r_{n}^{2} = \mu r_{n}^{2}\]

According to Boh's theory of equation of angular momentum,

\[\begin{array}{r} I\omega = \frac{nh}{2\pi}\ \Rightarrow \ \mu r_{n}^{2}\omega = \frac{nh}{2\pi}\#(iv) \end{array}\]

From equations (iii) and (iv), we get,

\[\begin{array}{r} r_{n} = \frac{\varepsilon_{0}n^{2}{\text{ }h}^{2}}{\pi\mu e^{2}Z}\#(v) \end{array}\]

Comparing this value with the value of \(r_{n}\) when nucleus was assumed be massive, we see that, ' m ' has been replaced by \(\mu\). Futher electrical potential energy of the system.

\[U_{n} = \frac{{Ze}^{2}}{4\pi\varepsilon_{0}r_{n}}\text{~}\text{and kinetic energy,}\text{~}K_{n} = \frac{1}{2}I\omega^{2} = \frac{1}{2}\mu r_{n}^{2}\omega^{2}\]

Explanation of bohr quantisation rule from de-Broglie's Concept

Einstein suggested that light behaves both as a material particle as well as wave. de-Brogle extended Einstein's view and said that all forms of matter like electrons, protons, neutrons etc. also dual character. He futher said that wavelength ( \(\lambda\) ) associated with a particle of mass ' \(m\) ' moving with velocity ' \(v\) ' is given by,

\[\lambda = \frac{h}{mv} = \frac{h}{p}\ \text{~}\text{(where}\text{~}\lambda\text{~}\text{is called de-Broglie's wavelength)}\text{~}\]

Futher : If the K.E. of the moving particle is K, then, \(\lambda = h/\sqrt{2mK}\)
If a charged particle ' \(q\) ' is accelerated through a potential difference \(\Delta V\) then, \(\lambda = h\sqrt{2mq(\Delta V)}\)
An electron behaves as standing or stationary wave, which extends round the nucleuses in a circular orbit. If the two ends of the electron wave meet to given a regular series of crests and troughs, the electron wave is said to be in phase. i.e. there is constructive interference of electron waves and the electron motion has a character of standing wave or non-energy radiation motion.
Watever be the path of the electron wave round the nucleus, it is a necessary condition to get an electron wave in phase so that the circumference of the Bohr's orbit \(( = 2\pi r)\) is equal to the whole number multiple to wavelength \(\lambda\) of electron wave i.e.

Circumference \(= 2\) wavelengths

Circumference \(= 4\) wavelengths

Circumference \(= 8\) wavelengths

Figure show some modes of vibration of a wire loop. In each case a whole number of wavelengths fit into the circumference of the loop.

\[\begin{array}{r} 2\pi r = n\lambda\text{~}\text{or}\text{~}\lambda = \frac{2\pi r}{n}\#(i) \end{array}\]

Where ' \(n\) ' is a whole number which denotes the number of wavelengths associated with an electron wave extending round the nucleus. Now according to de - Broglie,

\[\begin{array}{r} \lambda = \frac{h}{mv}\#(ii) \end{array}\]

From equation (i) and (ii) we get,

\[\frac{2\pi r}{n} = \frac{h}{mv}\text{~}\text{or}\text{~}mvr = \frac{nh}{2\pi}\]

Figure shows a fractional number of wavelengths cannot persist because destructive interference will occur.

An electron revolving in a permitted orbit does not radiate energy, through it is accelerating, so the total energy of the electron remains constant. That is why the permitted orbits are also called stationary or non-radiating orbits.

Energy of a trapped electron

The energy of a trapped particle is quantized.
The simplest case is that of a particle that bounces back and forth between the walls of a box. We shall assume that walls of the box are infinitely hard, so the particle does not lose energy each time it strikes a wall and that its velocity is sufficiently small so that we can ignore relativistic

A particle confined to a box of width L .

conditions.

From a wave points of view, a particle trapped in a box is like a standing wave in a string stretched between the box's walls. The possible de-Broglie wavelengths of the particle in the box therefore are determined by the width \(L\) of the box. The general formula for the permitted wavelength is given by.

\[\text{~}\text{de Broglie wavelengths of trapped particle,}\text{~}\lambda_{n} = \frac{2\text{ }L}{n},n = 1,2,3,\ldots\ldots.\]

Further for a matter wave de-Broglie wavelength,
\(\lambda_{n} = \frac{h}{{mv}_{n}} = \frac{h}{\sqrt{2{mK}_{n}}}\) (where \(k_{n}\) is the kinetic energy of the particle)

Comparing the two expressions, we have,

Wave functions of particle tapped in box of width L

\[\frac{h}{\sqrt{2{mk}_{n}}} = \frac{2\text{ }L}{n}\text{~}\text{or}\text{~}K_{n} = \frac{n^{2}{\text{ }h}^{2}}{8{\text{ }mL}^{2}}\]

Since the particle has no potential energy in this model, the only energies it can have, are,

Particle in a box, \(E_{n} = \frac{n^{2}{\text{ }h}^{2}}{8{\text{ }mL}^{2}},n = 1,2,3,\ldots\ldots\).

Each permitted energy is called an energy level and the integer ' \(n\) ' that specifies an energy level \(E_{n}\) is called its quantum number.
We can draw three general conclusions from the energy expression of a trapped electron. These conclusions apply to any particle confined to a certain region of space for instance, an atomic electron held captive by the attraction of the positively charge nucleus.

1. A trapped particle can not have an arbitrary energy, as free particle can have.

2. A trapped particle can not have zero energy.

3. Because Planck's constant is so small ( \(h = 6.63 \times 10^{- 34}\text{ }J\) sec), quantization of energy is conspicuous only when ' \(m\) ' and \(L\) are also small. That is why we are not aware of energy quantization in our own experience.

X-RAY

Historical background

On Nov, 1895, Wilhelm Conrad Röntgen (accidentally) discovered an image cast from his cathode ray generator, projected far beyond the possible range of the cathode rays (now known as an electron beam). Further investigation showed that the rays were generated at the point of contact of the cathode ray beam on the interior of the vacuum tube, that they were not deflected by magnetic fields, and they penetrated many kinds of matter. Röntgen named the new form of radiation X-radiation (X standing for "Unknown").

What are X-Rays

X-rays are electromagnetic radiation of very short wavelength ( \(0.1\text{Å}\) and \(100\text{Å}\) ) and high energy which are emitted when fast moving electrons or cathode rays strike a target of high atomic mass.

The X-Rays extends from the ultraviolet band to gamma rays band. Although their is very thin line between the X-ray and gamma rays but they are distinguished by the source of their generation. In the older days the X-Rays were distinguished from gamma rays by the wavelength and frequency but with time it has been observed that their exists X-Rays and gamma rays overlapping band and hence it is then understood that the distinction between them is source of generation.

Coolidge tube

William Coolidge invented the X -ray tube popularly called the Coolidge tube. His invention revolutionized the generation of X-rays and is the model upon which all X-ray tubes for medical applications are based.

The characteristic features of the Coolidge tube are its high vacuum and its use of a heated filament as the source of electrons. There is so little gas inside the tube that it is not involved in the production of x -rays, unlike the situation with cathode gas discharge tubes.
The operation of the Coolidge tube is as follows. Due to filament current the cathode filament is heated, it emits electrons. The hotter the filament gets, the greater the emission of electrons. A constant potential difference of several kilovolts is maintained between the filament and the target using a DC power supply so that the target is at a higher potential than the filament. These electrons are accelerated towards the anode with a very high speed and when the electrons strike the anode and emit x-rays. These X-rays are brought out of the tube through a window W made of thin mica or mylar or some such material which does not absorb X-rays appreciably. In the process, large amount of heat is developed, and thus an arrangement is provided to cool down the tube continuously by running water.

Note

(i)An increase in the filament current increase the number of electrons it emits. Larger number of electrons means an intense beam of X-rays is produced. This way we can control the quantity of X-rays i.e. Intensity of X-rays.
(ii) An increase in the voltage of the tube increase the kinetic energy of electrons \(\left( eV = 12{mv}^{2} \right)\). When such highly energetic beam of electrons are suddenly stooped by the target, an energetic beam of X-rays is produced. This way we can control the quality of X-rays i.e. penetration power of X-rays.
(iii) Based on penetrating power, X-rays are classified into two types. HARD-rays and SOFT-x-rays. The first one having high energy and hence high penetration power are HARD-X-rays and another one with low energy and hence low penetration power are SOFT-X-rays.

Continuous X-rays

When electron strikes target an electron loses a part of its kinetic energy and continues to move with the remaining energy until it hits another atom of the target. Part or whole of the energy lost by the electron is converted into a photon. This process is known as bremsstrahlung (breaking radiation)- as it leads to the electron getting decelerated by the target. There is existence of a minimum wavelength (or maximum frequency) in x-ray spectrum. This is called the cutoff wavelength or the threshold wavelength.

If the potential difference between the filament and the target is \(V\), then the kinetic energy of the electron just before it hits the target is

\[\text{~}\text{K.E.}\text{~} = eV\]

Energy of the X-ray photon, \(\Delta E = \frac{hc}{\lambda}\)
\(\Delta E\) is the fraction of \(K.E\). of electron that gets converted into photon. 1 is the wavelength of the photon. Wavelength of the X-ray's photon

\[\lambda = \frac{hc}{\Delta E}\]

as

\[\begin{matrix} \Delta E \leq eV & \Rightarrow \lambda \geq \frac{hc}{eV} \\ \Rightarrow & \lambda_{\min} = \frac{hc}{eV} \end{matrix}\]

This minimum wavelength does not depend on the target material and depend on the potential difference between the filament and the target.

Characteristic x-ray

When the high energy lectrons "knock off" the innermost electrons of the atoms of the target material causing a vacancy. This vacancy is filled by a electron that 'jump' from one of the outer shells. If the vacancy is created in K shell and electron from the L shell fills the vacancy then the emitted photon is a \(K_{\alpha}X\)-ray. If a vacancy is creasted in the K -shell and an electron from the M shell fills this vacancy then the \(x\)-ray emitted is known as a \(K_{\beta}X\)-ray.

\[\lambda_{K_{a}} = \frac{hc}{E_{L} - E_{K}}\]

\[\lambda_{K_{p}} = \frac{hc}{E_{M} - E_{K}}\]

Where \(E_{K},E_{L},E_{M}\) are the electron energy levels
These X-ray are known as characteristic X-ray as they depend on the target material (character of material), not on the applied voltage.

The adjoining graph shows the variation of the intensity of X -ray coming out of the tube with wavelengths. At some sharply defined wavelengths \(\left( K_{\alpha}K_{\beta} \right)\) the intensity of the emitted radiation is very large.

Moseley's Law

By measuring the wavelength associated with a particular line (now designated \(K_{\alpha}\) ), from the spectrum of each element, Moseley established that the lines from a large number of elements obeyed the relation

\[\sqrt{v} = a(Z - b)\]

Where a and b are constants with

\[a \approx \sqrt{\frac{3Rc}{4}}(R = \text{~}\text{Rydberg constant}\text{~})\]

and \(b \approx 1\). (b is known as the screening constant)

Approximate explanation from Bohr's theory

Consider an atom from which an electron from the K shell been knocked out. Consider an electron from the L shell which is about to make a transition to the vacant site. If finds the nucleus of charge Ze screened by the spherical cloud of the remaining one electron in the K shell. If we neglect the effect of the outer electrons and the other L electrons, the electron making the transition finds a charge \((Z - 1)e\) at the centre. One, therefore, may expect Bohr's model to given reasonable results if Z is replaced by \(Z - b\)

\[\begin{matrix} & \Delta E = hv = Rhc(Z - b)^{2}\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) \\ \Rightarrow \ & \sqrt{v} = \sqrt{\frac{3Rc}{4}}(Z - b) \end{matrix}\]

Properties of X-Ray

1. These are highly penetrating rays and can pass through several materials which are opaque to ordinary light.

2. They ionize the gas through which they pass. While passing through a gas, they knock out electrons from several of the neutral atoms, leaving these atoms with + ve charge.

3. They cause fluorescence in several materials. A plate coated with barium platinocyanide, ZnS ( zinc sulphide) etc becomes luminous when exposed to X-ray.

4. They affect photographic plates especially designed for the purpose.

5. They are not deflected by electric and magnetic fields, showing that they not charge particles.

Practice Exercise

Q. 1 If the operating potential in an X-ray tube is increased by \(1\%\), by what percentage does the cutoff wavelength decrease?
Q. 2 When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of \(9.7 \times 10^{18}\text{ }Hz\). Calculate the value of Planck constant from these data.
Q. 3 The \(K_{\beta}X\)-Ray of argon has a wavelength of 0.36 nm . The minimum energy needed to ionize an argon atom is 16 eV . Find the energy needed to knock out an electron from the K shell of an argon atom.
Q. 4 A free atom of iron emits \(K_{\alpha}X\)-rays of energy 6.4 keV . Calculate the recoil kinetic energy of the atom. Mass of an iron atom \(= 9.3 \times 10^{- 26}\text{ }kg\).

It exists at the centre of an atom, containing entire positive charge and almost whole of mass. The electron revolve around the nucleus to form an atom. The nucleus consists of protons (+ve charge) and neutrons. A proton has positive charge equal in magnitude to that of an electron ( \(+ 1.6 \times 10^{- 19}\) C ) and a mass equal to 1840 C ) and a mass equal to 1840 times that of an electron. A neutron has no charge and mass is approximately equal to that of proton.

Properties of a nucleus

(1) Nuclear Mass :

As we know that every nucleus contains protons and neutrons and so every nucleus has a definite mass. However, since the mass of electron is negligible so atomic mass is roughly equal to nuclear mass.

Atomic masses are measured in atomic mass unit (a.m.u.) defined as

\[\begin{matrix} & 1amu = 1.6604 \times 10^{- 27}\text{ }kg \\ \Rightarrow \ & 1u = 931.478MeV/c^{2} \end{matrix}\]

and its energy equivalent is 931.48 MeV
The number of protons in a nucleus of an atom is called as the atomic number \((Z)\) of that atom. The number of protons plus neutrons (called as Nucleus) in a nucleus of an atom is called as mass number (A) of that atom.

A particular set of nucleons forming an atom is called as nuclide. It is represented as \(\ _{Z}X^{A}\).The nuclides having same number of protons ( Z ), but different number of nucleons ( A ) are called as isotopes. The nuclide having same number of nucleons (A), but different number of protons (Z) are called as isobars. The nuclide having same number of neutrons (A-Z) are called as isotones.
(2) Nuclear charge

Since nucleus contain + vely charged protons ( charge \(= 1.6 \times 10^{- 19}C\) ) and neutrons (neutral) so every nucleus has a net + ve charge.
(3) Nuclear radius

A rough estimate of nuclear size suggests us that the radius of the nucleus of an atom having mass number ' A ' is given by

\[R = R_{0}{\text{ }A}^{1/3}\]

Where \(R_{0}\) is a constant found to be equal to

\[R_{0} = 1.4 \times 10^{- 15}\text{ }m = 1.4fm\]

(4) Nuclear Density

In spite of the fact that nuclear radius depends on mass number of the atom but nuclear density is independent of mass number because if neutrons are supposed to be of almost the same mass as that of protons then the total mass of a nucleus is proportional to A . If each nucleon are
supposed to have a mass \(m\) then nuclear density is given by

\[\rho = \frac{mA}{\frac{4}{3}\pi R_{0}^{3}\text{ }A} = \frac{3\text{ }m}{4\pi R_{0}^{3}}\]

(Which is independent of A)

(5) Nuclear spin and magnetic moment

Like orbital electrons in an atom, nucleons inside nucleus have well defined quantum states. Correspondingly they have angular momentum and hence a magnetic moment. Like electrons nucleons also have intrinsic angular momentum and 'magnetic' moment corresponding to their spin.

Nuclear Forces

If only the electrostatic and gravitational forces existed in the nucleus, then it would be impossible to have stable nuclei composed of protons and neutrons. The gravitational forces are much too small to hold the nucleons together compared to the electrostatic forces repelling the protons. Since stable atoms of neutrons and protons do exist, there must be another attractive force acting within the nucleus. This force is called the nuclear force.

Properties of nuclear force

(1) They are charge independent. The nuclear force between two proton is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces.
(2) They may be repulsive may be attractive (Repulsive at exceedingly small separation between two nucleons appreciably smaller than \(10^{- 13}\text{ }cm\) i.e. \(10^{- 15}\text{ }m\)
(3) It is a short range force. Its radius of action is of the order of \(10^{- 13}\text{ }cm\).
(4) The nuclear force is of saturation character. Each nucleon in nucleus interacts with a limited number of nucleons.
(5) Nuclear force are much stronger than electromagnetic force or gravitational attractive forces. It is the strongest of all the forces. This is why it is called strong interaction.
(6) Nuclear force is spin dependent. If two interacting nucleons are having parallel spins then nuclear force operative between them is comparatively stronger and if their spins are antiparallel, nuclear interaction is comparatively weaker.
(7) Nuclear force is a non-central force. They can not be represented as directed along the st. line connecting the centres of the interacting nucleons. Its non central nature is due to the fact that it depends also on the orientation of the nucleon spins.

Mass defect

It is observed that the mass of a nucleus is slightly less than the sum of the masses of constituent nucleons. Suppose a nucleus consists of ' \(Z\) ' protons and ' \(N\) ' neutrons. Mass of a proton, a neutron and the resulting nucleus are respectively \(m_{p},m_{n}\) and \(M\) then mass defect of the nucleus is given by

\[\Delta m = Zm_{p} + Nm_{n} - M\]

If A is the mass number of the nucleus

\[\Delta m = \left\lbrack {Zm}_{p} + (A - Z)m_{n} - M \right\rbrack\]

In terms of atomic masses we may also write mass defect as

\[\Delta m = \left\lbrack Zm\left( \ _{1}^{1}H \right) + {Nm}_{n} - m\left( \ _{z}^{A}X \right) \right\rbrack\]

Where \(m\left( \ _{1}^{1}H \right) =\) mass of one hydrogen atom.
\(m\left( \ _{z}^{A}X \right) =\) mass of atom having atomic no. Z and mass no. A
e.g.,
mass of \(\ _{1}^{1}H = 1.00784u\)
mass of neutron \(= 1.00874u\)
Expected mass of deuterium \(= 2.01654u\)
but measured mass \(= 2.0141u\).
mass defect,

\[\Delta m = 0.00244u.\]

Nuclear Stability

Figure shows pot of N vs. Z for known nuclides. The stable nuclides are indicated by the black dots. Non-stable nuclides decay by emission of particles, or electromagnetic radiation, in a process called radioactivity

Binding energy

To break a nucleus into its constituent nuclei some energy is required to be supplied. This energy is called Binding Energy of the given nucleus or the energy equivalent of the missing mass of a nucleus is called the binding energy of the nucleus.
\(BE = (\Delta m)c^{2} = \left\lbrack {Zm}_{p} + (A - Z)m_{n} - M \right\rbrack c^{2}\),
\(BE = \Delta m(\) in amu \() \times 931MeV\ \) Where \(\Delta m =\) mass defect
Binding energy per nucleon is a measure of the stability of the nucleus. If there be n nucleons which is equal to A ,
\[\frac{\text{~}\text{Binding Energy}\text{~}}{\text{~}\text{Nucleon}\text{~}} = \frac{\text{~}\text{B.E.}\text{~}}{A} \]

From the plot of B.E. / nucleons Vs mass number (A), we observe that :

(1) Binding energy per nucleon has low value for both heavy and light nuclei i.e. Heavy as well as light nuclei, both are unstable.B.E. / nucleons increases on an average and reaches a maximum of about 8.7 MeV for \(A \equiv 50 - 80\). For more heavy nuclei, B.E. / nucleons decreases slowly as A increases. For the heaviest natural element \(U^{238}\) it drops to about 7.5 MeV . From above observation, if follows that nuclei in the region of atomic masses \(50 - 80\) are most stable.
(2) The intermediate nuclei have large value of binding energy per nucleon so they are more stable.
(3) Binding energy per nucleon increases rapidly upto mass number 20 but there are peaks corresponding to \(\ _{2}^{4}He,\ _{6}^{12}C,\ _{8}^{16}O\) which indicates that these nuclei are more stable than neighbours. The reason is that they may be considered to posses magic numbers i.e. their mass number is divisible by 4 and these nuclei may have \(\ _{2}^{4}He\) as their constituents.
(4) The minimum value of the BE/Nucleon is in the case of deuteron that is 1.11 Mev .
(5) The maximum value of the BE/Nucleon is 8.79 Mev for the nuclide \(\ _{26}^{56}Fe\) which is therefore the most stable nucleus.

Note

(i) If we split a heavy nucleus into two medium sized nuclei and total binding energy of new nuclei is greater then parent nuclei, then energy is released (Nuclear fission)
(ii) If two nuclei of small mass number combine to form a single medium size nucleus for which binding energy is greater than the constituent nuclei, then energy is released (Nuclear fusion)

Nuclear Fission

The breaking of a heavy nucleus into two or more fragments of comparable mass, with the release of tremendous energy is called as nuclear fission.

The most typical fission reaction occurs when slow moving neutrons strike \(\ _{92}U^{235}\). The following nuclear reaction takes place.

\[\ _{92}U^{235} + \ _{0}n^{1} \longrightarrow \ _{56}{Ba}^{141} + \ _{36}{Kr}^{92} + 30_{0}n^{1} + 200MeV\]

If more than one of the neutrons produced in the above fission reaction are capable of inducing a fission reaction (provided \(U^{235}\) is available), then the number of fission taking place at successive stages goes increasing at a very brisk rate and this generates a series of fission. This is known as chain reaction. The chain reaction takes place only if the size of the fissionable material ( \(U^{235}\) ) is greater than a certain size called the critical size.

If the number of fission in a given interval of time goes on increasing continuously, then a condition of explosion is created. In such cases, the chain reaction is known as uncontrolled chain reaction. This forms the basis of atomic bomb.

In a chain reaction, the fast moving neutrons are absorbed by certain substances known as moderators (like heavy water), then the number of fissions can be controlled and the chain reaction is such cases is known as controlled chain reaction. This forms the basis of a nuclear reactor.

Nuclear Fusion

The process in which two or more light nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear fusion. Like a fission reaction, the sum of masses before the fusion (i.e. of bigger nucleus) and this difference appears as the fusion energy. The most typical fusion reaction is the fusion of two deuterium nuclei into helium.

\[\ _{1}H^{2} + \ _{1}H^{2} \longrightarrow \ _{2}{He}^{4} + 21.6MeV\]

For the fusion reaction to occur, the light nuclei are brought closer to each other (with a distance of \(10^{- 14}\text{ }m\) ). This is possible only at very high temperature to counter the repulsive force between nuclei. Due to this reason, the fusion reaction is very difficult to perform. The inner core of sun is at very high temperature, and is suitable for fusion. In fact the source of sun's and other star's energy is the nuclear fusion reaction.

Conservation laws in nuclear reaction

Nuclear reaction processes have led to the formulation of useful conservation principles. The four principles of most interest in this module are discussed below.
(i) Conservation of electric charge implies that charges are neither created nor destroyed. Single positive and negative charges may, however, neutralize each other. It is also possible for a neutral particle to produce one charge of each sign.
(ii) Conservation of mass number does not allow a net change in the number of nucleons i.e. total number of protons and neutrons should also remain same on both sides of a nuclear reaction.. However, the conversion of a proton to a neutron and vice versa is allowed.
(iii) Conservation of mass and energy implies that the total of the kinetic energy and the energy equivalent of the mass in a system must be conserved in all decays and reactions. Mass can be converted to energy and energy can be converted to mass, but the sum of mass and energy must be constant. In nuclear reactions, sum of masses before reaction is greater than the sum of masses after the reaction. The difference in masses appears in form of energy following the Law of inter-conversion of mass & energy. The energy released in a nuclear reaction is called as Q Value of a reaction and is given as follows.
If difference in mass before and after the reaction is \(\Delta m\) amu ( \(\Delta m =\) mass of reactants minus mass of products) then

\[Q\text{~}\text{value}\text{~} = \Delta m(931)MeV\]

(iv) Conservation of momentum is responsible for the distribution of the available kinetic energy among product nuclei, particles, and/or radiation. The total amount is the same before and after the reaction even though it may be distributed differently among entirely different nuclides and/or particles.

Pratice Exercise

Q. 1 Calculate the electric potential energy due to the electric repulsion between two nuclei of \(\ ^{12}C\) when they 'touch' each other at the surface.
Q. 2 Find the binding energy of \(\ _{26}{Fe}^{56}\). Atomic mass of \(\ _{26}{Fe}^{56}\) is 55.9349 u and that of \(\ ^{1}H\) is 1.00783 u . Mass of neutron is \(1 \cdot 00867u\).
Q. 3 Calculate the Q-value in the following decay
\[\ ^{25}Al \rightarrow \ ^{25}Mg + e^{+} + v \]Q. 4 Find the maximum energy that a beta particle can have in the following decay

\[\ ^{176}Lu \rightarrow \ ^{176}Hf + e + \overline{v}\]

Atomic mass of \(\ ^{176}Lu\) is 175.942694 u and that of \(\ ^{176}Hf\) is 175.941420 u

Q. 1 A point source of light is placed at the centre of curvature of a hemispherical surface. The radius of curvature is \(r\) and the inner surface is completely reflecting. Find the force on the hemisphere due to the light falling on it if the source emits a power W.
Sol. The energy emitted by the source per unit time, i.e. W fall on a area \(4\pi r^{2}\) at a distance r in unit time. Thus, the energy falling per unit area per unit time is \(\frac{W}{4\pi\pi^{2}}\). Consider a small area dA at the point P of the hemisphere (figure).
The energy falling per unit time on it, is

\[P = \frac{WdA}{4\pi r^{2}}\]

The corresponding momentum incident on this area per unit time is

\[= \frac{WdA}{4\pi r^{2}c}\]

Suppose the radius OP through the area dA makes an angle \(\theta\) with the symmetry axis OX . The force on dA is along this radius.

\[dF = \frac{2WdA}{4\pi r^{2}c}\]

By symmetry, the resultant force on the hemisphere is along OX . The component of dF along OX is

\[\begin{matrix} & dFcos\theta = \frac{2WdA}{4\pi r^{2}c}cos\theta \\ & \ = \frac{2\text{ }W}{4\pi r^{2}c}\text{~}\text{(projection the area}\text{~}dA\text{~}\text{on the plane containing the rim)}\text{~} \end{matrix}\]

The net force along OX is

\[\begin{matrix} F & \ = \frac{2\text{ }W}{4\pi r^{2}c}\Sigma(\text{~}\text{projec}\text{tion the area}\text{~}dA\text{~}\text{on the plane containing the rim}\text{~}) \\ & \ = \frac{2\text{ }W}{4\pi r^{2}c}\left( \pi r^{2} \right) = \frac{W}{2c} \end{matrix}\]

Q. 2 Find the maximum kinetic energy of photo-electron liberated from the surface of lithium ( \(\phi = 2.39eV\) ) by electromagnetic radiation whose electric component varies with time as \(E = a(1 + cos\omega t)cos\omega_{0}I\), where ' \(a\) ' is a constant. \(\omega = 6 \times 10^{14}rad/sec\) and \(\omega_{0} = 3.60 \times 10^{15}rad/s\).
Sol. \(\ E = a(1 + cos\omega t)cos\omega_{0}t = acos\omega_{0}t = cos\omega_{0}t + acos\omega tcos\omega_{0}t\)
\[\Rightarrow E = acos\omega_{0}t + \frac{1}{2}acos\left( \omega + \omega_{0} \right)t + \frac{1}{2}acos\left( \omega + \omega_{0} \right)t \]This is a complex vibration consisting of harmonic components of frequencies \(\omega_{0},\left( \omega + \omega_{0} \right)\) and \(\left( \omega - \omega_{0} \right)\).

The highest angular frequency is \(\left( \omega + \omega_{0} \right)\).
Now, \(h\nu = \phi + k_{\text{max}\text{~}}\)
So, \(k_{\max} = \frac{h}{2\pi}\left( \omega + \omega_{0} \right) - \phi\)
\[{= \frac{6.6 \times 10^{- 34}}{2\pi}\left( 6 \times 10^{14} + 3.6 \times 10^{15} \right) - 2.39 \times 1.6 \times 10^{- 19} }{= 4.41 \times 10^{- 19} - 3.82 \times 10^{- 19} = 0.59 \times 10^{- 19}\text{ }J = 0.37eV }\]Q. 3 Find the ratio of de-Broglie wavelength of an \(\alpha\)-particle to that of a proton being subjected to the same magnetic field so that the radii of their paths are equal to each other, assuming that the field induction vector \(\overrightarrow{B}\) is perpendicular to the velocity vectors of the a-particle and the proton.
Sol. When a charged particle of charge \(q\), mass \(m\) enters perpendicularly to the magnetic induction \(\overrightarrow{B}\) of a magnetic field, it will experience a magnetic force
\(F = q(\overrightarrow{v} \times \overrightarrow{B}) = qvBsin90^{\circ} = qvB\) that will provide a centripetal acceleration \(\frac{v^{2}}{r}\)
\[{\Rightarrow qvB = \frac{{mv}^{2}}{r} }{\Rightarrow mv = qBr \Rightarrow}\] The de-Broglie wavelength \(\lambda = \frac{h}{mv} = \frac{h}{qBr}\)
\[\Rightarrow \frac{\lambda_{\alpha - \text{~}\text{particle}\text{~}}}{\lambda_{\text{proton}\text{~}}} = \frac{q_{p}r_{p}}{q_{\alpha}r_{\alpha}} \]Since \(\frac{r_{\alpha}}{r_{p}} = I\) and \(\frac{q_{\alpha}}{q_{p}} = 2\)
\[\Rightarrow \frac{\lambda_{\alpha}}{\lambda_{p}} = 1/2 \]Q. 4 A particle of mass \(m\) moves along a circular orbit in a centrosymmetric potential field \(U = \frac{{kr}^{2}}{2}\). Using Bohr's quantization condition. Find (a) radius of \(n^{\text{th}\text{~}}\) orbit (b) Energy of \(n^{\text{th}\text{~}}\) orbit

Sol.

\[F = - \frac{dU}{dr} = - kr\]

so \(\ kr = \frac{{mv}^{2}}{r}\)
and \(\ mvr = \frac{nh}{2\pi}\)

Solving we get \(r = \left\lbrack \frac{h^{2}n^{2}}{4\pi mk} \right\rbrack^{\frac{1}{2}}\)
Total energy \(E_{n} = {KE}_{n} + {PE}_{n}\)

\[\begin{matrix} & \ = \frac{1}{2}{mv}^{2} + \frac{{kr}^{2}}{2} = {kr}^{2} \\ & \ = K\sqrt{\frac{n^{2}{\text{ }h}^{2}}{4\pi^{2}mK}} = \sqrt{\frac{n^{2}{\text{ }h}^{2}{\text{ }K}^{2}}{4\pi^{2}mK}} \\ & \ = \frac{nhK}{2\pi\sqrt{mK}} \end{matrix}\]

Q. 5 Compare the radii and energy of ground state of H -atom and p -atom considering the motion of nucleus.

Sol. If we consider the motion of nucleus mass of \(e^{-}\)in all the expressions will be replaced by \(\mu\). Where

\[\mu_{H} = \frac{mM}{\text{ }m + M};\ M = \text{~}\text{Ma}\text{ss of Proton or neutron}\text{~}\]

and

\[\mu_{D} = \frac{m(2M)}{m + 2M}\]

hence

\[\mu_{D} > \mu_{H}\]

radius of \(n^{\text{th}\text{~}}\) orbit \(r_{n} \propto \frac{1}{\mu}\) so \(r_{H} > r_{D}\)
Energy of \(n^{\text{th}\text{~}}\) orbit \(E_{n} \propto \mu\) so \(E_{\mu} < E_{D}\).
Q. 6 What lines of atomic hydrogen absorbing spectrum fall with in the ware length range from 94.5 nm to 130 nm .
Sol. Absorbtion lines are always corresponding to lyman series.
Wave length of lyman series \(\frac{1}{\lambda} = R\left\lbrack \frac{1}{12} - \frac{1}{n^{2}} \right\rbrack\)
for \(n = 2,\lambda_{1} = 121\text{ }nm\)
for \(n = 3,\lambda_{2} = 102.2\text{ }nm\)
for \(n = 4,\lambda_{3} = 96.9\text{ }nm\)
for \(n = 5,\lambda_{4} = 94.64\text{ }nm\)
for \(n = 6,\lambda_{5} = 93.45\text{ }nm\)
hence 121.1 nm 102.2 nm 96.9 nm and 94.64 nm .
Q. 7 A stationary H -atom emits a photon corresponding to the friest line of lyman series. What velocity does the atom acquire? \(\left( M_{H} = 1.67 \times 10^{- 19}\text{ }kg \right)\)

Sol.

Applying momentum and energy conservation,

\[mv = \frac{hv}{C}\text{~}\text{and}\text{~}\Delta E = \frac{1}{2}{mv}^{2} + hv\]

when \(\Delta E = 10.2 \times 1.6 \times 10^{- 19}\) Joule
we get \(\Delta E = \frac{1}{2}{mv}^{2} + mvC\)

\[\begin{matrix} & \Delta E = mvC\left\lbrack \frac{\text{ }V}{2C} + 1 \right\rbrack \approx mvC \\ & v = \frac{DV}{mC} = \frac{10.2 \times 1.6 \times 10^{- 19}}{1.6 \times 10^{- 27} \times 3 \times 10^{8}} \\ & \ = 3.25\text{ }m/sec \end{matrix}\]

Q. 8 The BE of an electron in ground state of the atom is equal to \(E_{0} = 24.6ev\). Find the energy required to remove both electron from the atom.
Sol Ionisation energy of \({He}^{+}\)atom \(= 54.4ev\) hence to remove both electrons from He -atom we require \(= 24.6 + 54.4 = 79ev\)
Q. 9 An X-ray tube with a copper target is found the emit lines other than those due to copper. The \(k_{a}\) line of copper is known to have a wavelength \(1.5405\text{Å}\) and the other two \(K_{a}\) lines observed have wavelengths \(0.7092\text{Å}\) and \(1.6578\text{Å}\). (Identify) the impurities (find the value of Z , atomic number).
Sol. According to Moseley's equation for \(K_{a}\) radiation

\[\frac{1}{\lambda} = R(Z - 1)^{2}\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack\text{~}\text{where}\text{~}\lambda = \text{~}\text{wavelength of copper}\text{~}\]

Let \(\lambda_{1}\) and \(\lambda_{2}\) be the two other unknown wavelengths, then

\[\frac{\lambda_{1}}{\lambda} = \frac{(Z - 1)^{2}}{\left( Z_{1} - 1 \right)^{2}} = \frac{0.7092}{1.5405}\]

Solving we get \(Z_{1} = 42\)
Similarly

\[\frac{\lambda_{2}}{\lambda} = \frac{(28)^{2}}{\left( Z_{2} - 1 \right)^{2}} = \frac{1.6578}{1.5405}\]

Solving we get \(Z_{2} = 28\)
Q. 10 When \(0.50\text{Å}X\)-rays strike a material, the electrons from the k shell are observed to move in a circle of radius 23 mm in a magnetic field of \(2 \times 10^{- 2}\text{ }T\). What is the binding energy of K -shell electrons?
Sol. The velocity of the photoelectrons is found the \(F = ma\) :
\(evB = m\frac{v^{2}}{R}\ \) or \(\ v = \frac{e}{m}BR\)
The kinetic energy of the photoelctrons is then
\[{K = \frac{1}{2}{mv}^{2} = \frac{1}{2}\frac{e^{2}{\text{ }B}^{2}R^{2}}{\text{ }m} }{= \frac{1}{2}\frac{\left( {1.6510}^{- 19}C \right)^{2}\left( 2 \times 10^{- 2}\text{ }T \right)^{2}\left( 23 \times 10^{- 3}\text{ }m \right)^{2}}{\left( 9.1 \times 10^{- 31}\text{ }kg \right)} = 2.97 \times 10^{- 15}\text{ }J }\]or \(K = \left( 2.97 \times 10^{- 15}\text{ }J \right)\frac{1keV}{1.6 \times 10^{- 16}\text{ }J} = 18.36eV\)
The energy of the incident photon is \(E_{v} = \frac{hc}{\lambda} = \frac{12.4keV.\text{Å}}{0.50\text{Å}} = 24.8eV\)
The binding energy is the difference between these two value:

\[BE = Ev - K = 24keV - 18.6keV = 6.2keV\]

Q. 11 Calculate the wavelength of the emitted characteristic X-ray from a tungsten ( \(Z = 74\) ) target when an electron drops from an \(M\) shall to a vacancy in the \(K\) shell.
Sol. Tungsten is a multiel atom. Due to the shielding of the nuclear charge by the negative charge of the inner core electrons, each electron is subject to an effective nuclaer charge \(Z_{\text{eff}\text{~}}\) which is different for different shells.
Thus, the energy of an electron in the \(n^{\text{th}\text{~}}\) level of a multielectron atom is given by

\[E_{n} = \frac{13.6Z_{eff}^{2}}{n^{2}}eV\]

For an electron in the K shell \((n = 1),Z_{\text{eff}\text{~}} = (Z - 1)\).
Thus, the energy of the electron in the K shell is :

\[E_{K} = - \frac{(74 - 1)^{2} \times 13.6}{1^{2}} \simeq - 72500eV\]

For an electron in the \(M\) shell \((n = 3)\), the nucleus is shielded by one electron of the \(n = 1\) state and eight electrons of the \(n = 2\) state, a total of nine electrons, so that \(Z_{eff} = Z - 9\). Thus the energy of an electron in the \(M\) shell is :

\[E_{M} = \frac{(74 - 9)^{2} \times 13.6}{3^{2}} \simeq - 6380eV\]

Therefore, the emitted X-ray photon has an energy given by

\[hv = E_{M} - E_{K} = - 6380eV - ( - 72500eV) = 66100eV\]

or \(\ \frac{hc}{\lambda} = 66100 \times 1.6 \times 10^{- 19}\text{ }J\)
\[{\therefore\ \lambda = \frac{hc}{66100 \times 1.6 \times 10^{- 19}}\text{ }m }{= \frac{\left( 6.63 \times 10^{- 34} \right) \times \left( 3 \times 10^{8} \right)}{66100 \times 1.6 \times 10^{- 19}}\text{ }m = 0.0188 \times 10^{- 9}\text{ }m}\].
Q. 12 If the short series limit of the Balmer series for hydrogen is \(3646\text{Å}\), calculate the atomic no. of the element which gives X -ray wavelength down to 1.0 A . Identify the element.
Sol. The short limit of the Balmer series is given by

\[\begin{matrix} & \overline{v} = 1/\lambda = R\left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right) = R/4 \\ \therefore & R = 4/\lambda = (4/3646) \times 10^{10}{\text{ }m}^{- 1} \end{matrix}\]

Futher the wavelengths of the \(k_{0}\) series are given by the relation

\[\overline{v} = \frac{1}{\lambda} = R(Z - 1)^{2}\]

or \(\ (Z - 1)^{2} = \frac{1}{R\lambda} = \frac{3646 \times 10^{- 10}}{4 \times 1 \times 10^{- 10}} = 911.5\)
\[{\therefore\ (Z - 1) = \sqrt{911.5} }{\cong 30.2 }\]or \(\ Z = 30.2 \cong 31\)
Thus the atomic number of the element concerned is 31 .
The element having atomic number \(Z = 31\) is Gallium.
Q. 13 In a nuclear reactor, fission is produced in 1 gm for \(U^{235}\) (235.0439 a.m.u.) in 24 hours by a slow neutron (1.0087 a.m.u.). Assuming that \(\ _{35}{Kr}^{92}\) (91.8973 a.m.u.) and \(\ _{56}{Ba}^{141}\) (140.9139 a.m.u.) are produced in all reactions and no energy is lost, write the complete reaction and calculate the total energy produced in kilowatt hour. Given 1 a.m.u. \(= 931MeV\).

Sol. The nuclear fission reaction is

\[\ _{92}U^{235} + \ _{0}n^{1} \rightarrow \ _{56}{Ba}^{141} + \ _{36}{Kr}^{92} + 3\ _{0}n^{1}\]

The sum of the masses before reaction

\[= 235.0439 + 1.0087 = 236.0526\text{~}\text{a.m.u.}\text{~}\]

The sum of the masses after reaction

\[\begin{matrix} & \ = 140.9139 + 91.8973 + (1.0087) = 235.8375\text{~}\text{a.m.u.}\text{~} \\ \Rightarrow & \Delta m = 256.0526 - 235.8373 = 0.2153\text{~}\text{a.m.u.}\text{~} \end{matrix}\]

energy released in the fission of \(U^{235}\) nucleus

\[E = 0.2153 \times 931 = 200Mev\]

Number of atoms in 1 gm

\[= \frac{6.02 \times 10^{23}}{235} = 2.56 \times 10^{21}\]

Energy released in fission of 1 gm of \(U^{235}\)

\[\begin{matrix} E & \ = 200 \times 2.56 \times 10^{2} = 5.12 \times 10^{23}MeV \\ & \ = \left( 5.12 \times 10^{23} \right) \times \left( 1.6 \times 10^{- 13} \right) = 8.2 \times 10^{10}\text{~}\text{joule}\text{~} \\ & \ = \frac{8.2 \times 10^{10}}{3.6 \times 10^{6}}kWh = 2.28 \times 10^{4}kWh \end{matrix}\]

Q. 14 A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron (a) in a head-on collision; (b) in scattering at right angles.

Sol. (a) In a heat on collision \(\sqrt{2mK} = p_{d} + p_{n}\)

\[K = \frac{p_{d}^{2}}{2M} + \frac{p_{n}^{2}}{2\text{ }m}\]

where \(p_{d}\) and \(p_{n}\) are the momenta of deuteron and neutron after the collision. Squaring

\[\begin{matrix} & p_{d}^{2} + p_{n}^{2} + 2p_{d}p_{n} = 2mK \\ & p_{n}^{2} + \frac{m}{M}p_{d}^{2} = 2mK \end{matrix}\]

or since \(p_{d} \neq 0\) in a head on collision

\[p_{n} = - \frac{1}{2}\left( 1 - \frac{m}{M} \right)P_{d}\]

Going back to energy conservation

\[\frac{p_{d}^{2}}{2M}\left\lbrack 1 + \frac{M}{4\text{ }m}\left( 1 - \frac{m}{M} \right)^{2} \right\rbrack = K\]

So \(\ \frac{p_{d}^{2}}{2M} = \frac{4mM}{(m + M)^{2}}K\)
This is the energy lost by neutron. So the fraction of energy lost is

\[\eta = \frac{4mM}{(\text{ }m + M)^{2}} = \frac{8}{9}\]

(b) In this case neutron is scattered by \(90^{\circ}\). Then we have from the diagram

\[{\overrightarrow{p}}_{d} = p_{n}\widehat{j} + \sqrt{2mK}\widehat{i}\]

Then by energy conservation

\[\begin{matrix} & \frac{p_{n}^{2} + 2mK}{2M} + \frac{p_{n}^{2}}{2m} = K \\ \text{~}\text{or}\text{~} & \frac{p_{n}^{2}}{2m}\left( 1 + \frac{m}{M} \right) = K\left( 1 - \frac{m}{M} \right) \\ \text{~}\text{or}\text{~} & \frac{p_{n}^{2}}{2m}\left( 1 + \frac{m}{M} \right) = K\left( 1 - \frac{m}{M} \right) \end{matrix}\]

The energy lost by neutron in then

\[K - \frac{p_{n}^{2}}{2\text{ }m} = \frac{2\text{ }m}{M + m}\text{ }K\]

or fraction of energy lost is

\[\eta = \frac{2\text{ }m}{M + m} = \frac{2}{3}\]

Q. 15 A stationary \({Pb}^{200}\) nucleus emits an \(\alpha\)-particle with \(K.E.,K = 5.77MeV\). Find the recoil velocity of daughter nucleus. What fraction of the total energy liberated in this decay is accounted for the recoil energy of the daughter nucleus?
Sol. The momentum of the a-particle is given by,

\[\begin{array}{r} P_{d} = P_{\alpha} = \sqrt{2{\text{ }m}_{\alpha}K}\#(i) \end{array}\]

Let the recoiled momentum of the dauhter nucleus be \(P_{d} = m_{d}v_{d}\), where \(m_{d}\) and \(v_{d}\) are the mass and velocity of daughter nucles. Using the priciple of conservation of momentum we get,

\[\begin{matrix} & & P_{d} = P_{\alpha} = \sqrt{2m_{\alpha}K} \\ \Rightarrow & V_{d} = \frac{\sqrt{2m_{\alpha}K}}{m_{d}} & \text{(ii)} \\ \Rightarrow & V_{d} = \frac{1}{196}\sqrt{\frac{2 \times 4 \times K}{m_{p}}} = \frac{2}{196}\sqrt{\frac{2K}{m_{p}}} & \text{(ii)} \end{matrix}\]

Where \(m_{p}\) is the mass of the proton.
\[\Rightarrow \ V_{d} = 3.39 \times 10^{5}\text{ }m/s \]Let the K.E. of the daughter nucleus be \(K^{'}\) then,

\[\frac{K^{'}}{K} = \frac{m_{\alpha}}{m_{d}}\]

As the momenta are same
\[{\therefore\ \frac{K^{'}}{K_{t}} = \frac{m_{\alpha}}{m_{\alpha} + m_{d}} }{\Rightarrow \ K^{'} = \frac{m_{\alpha}}{m_{\alpha} + m_{d}}K_{t} = \frac{4}{196 + 4}{\text{ }K}_{t} }{\Rightarrow \ K^{'} = 0.02{\text{ }K}_{t} }{\Rightarrow \ \frac{K^{'}}{K_{t}} = 0.02 }\]Q. \(16{AP}^{32}\) radionuclide with half-life \(T = 14.3\) days is produced in a reactor at a constant rate \(q = 2.7 \times 10^{9}\) nuclei per second. How soon after the beginning of production of that radionuclide will its activity be equal to \(A = 1.0 \times 10^{9}dps\) ?

Sol. Production of the nucleus is governed by the equation

\[\frac{dN}{dt} = \overset{\overset{g}{\uparrow} - \lambda\overset{t}{\uparrow} - \lambda\overset{N}{\uparrow}}{\uparrow}{decay}^{- \lambda\underset{d}{\text{ }N}}\]

we see that N will approach a constant value \(\frac{g}{\lambda}\). This can also be proved directly. Multiply by \(e^{\lambda t}\) and write.

\[\frac{dN}{dt}e^{\lambda,t} + \lambda e^{\lambda,t}\text{ }N = {ge}^{\lambda t}\]

Then

\[\frac{d}{dt}\left( {Ne}^{\lambda t} \right) = {ge}^{\lambda t}\]

or \(\ {Ne}^{\lambda t} = \frac{g}{\lambda}e^{\lambda t} +\) const.
At \(t = 0\) when the production is starteed, \(N = 0\)

\[0 = \frac{g}{\lambda} + \text{~}\text{constant}\text{~}\]

Hence

\[N = \frac{g}{\lambda}\left( 1 - e^{- \lambda t} \right)\]

Now the activity is

\[A = \lambda N = g\left( 1 - e^{- \lambda t} \right)\]

From the problem

\[\frac{1}{2 \cdot 7} = 1 - e^{- \lambda t}\]

This gives \(\lambda t = 0.463\)
so

\[t = \frac{0.463}{\lambda} = \frac{0.463 \times T}{0.693} = 9.5\text{~}\text{days.}\text{~}\]

Algebraically \(\ t = - \frac{T}{ln2}ln\left( 1 - \frac{A}{g} \right)\)
Q. 17 A dose of 5 mCi of \(P^{32}\left( t_{1/2} = 14 \right.\ \) days \()\) is a administered intravenously to a patient whose blood volume is \(3.5\mathcal{l}\) ts. At the end of 1 hour, it is assumed that the phosphorous is uniformly distributed.
What would be the count rate/m \(\mathcal{l}\) of the withdrawn blood if the counter measuring the activity had an efficiency of \(10\%\) :
(a) 1 hour after injection
(b) 28 days after injection

Sol. Let \(A_{0} =\) initial activity \(A_{t} =\) activity at time t
According to question

\[A_{0} = \frac{5 \times 10^{- 5}}{35 \times 10^{3}} = 0.143 \times 10^{- 5}Ci/m\mathcal{l}\]

and

\[\lambda = \frac{0.693}{14 \times 24} = 2.06 \times 10^{- 3}/Hr\]

Now using

\[\lambda t = 2.303log\frac{{\text{ }A}_{0}}{{\text{ }A}_{t}}\]

(a) After 1.0 Hr

\[\frac{2.06 \times 10^{- 3} \times 1}{2.303} = log\frac{0.143 \times 10^{- 5}}{A_{t}}\]

\[\Rightarrow \ A_{t} = 1.42 \times 10^{- 6}Ci/m\mathcal{l} \Rightarrow \ \] Count rate \(= (10/100) \times \left( 1.42 \times 10^{- 6} \right) \times 3.7 \times 10^{10}dps = 5280dps\)
(b) After 28 days, i.e., after two half lives \(\left( t_{1/2} = \right.\ \) of \(P^{32} = 14\) days \()\);
\[A_{t} = A_{0}/4 = 1.42 \times 10^{- 6}/4 \Rightarrow \ \] Count rate \(= (10/100) \times \left( 1.42 \times 10^{- 6}/4 \right) \times 3.7 \times 10^{10}dps = 1322.75dps\)
Q. 18 In the chemical analysis of a rock, the mass ratio of two radioactive isotopes is found to be \(100:1\). The mean lives of the two isotopes are \(4 \times 10^{9}\) and \(2 \times 10^{9}\) years respectively. If it is assumed that at the time of formation the atoms of both the two isotopes were in equal proportion, calculate the age of the rock. Ratio of the atomic weights of two isotopes is \(1.02:1\).
Sol. Let two isotopes are A and B
\[\frac{m_{A}}{m_{B}} = 100;\frac{A_{A}}{A_{B}} = \frac{1.02}{1} T_{A} = 4 \times 10^{9}\] years \(T_{B} = 2 \times 109\) years \(\ \lbrack\) Also \(\lambda = 1/T\rbrack\)
Let ratio of nuclei of two isotops be :

\[\frac{N_{A0}}{{\text{ }N}_{B0}}\text{~}\text{at}\text{~}t = 0\text{~}\text{and}\text{~}\frac{N_{At}}{{\text{ }N}_{Bt}}\text{~}\text{at}\text{~}t = 1\]

For isotope A

\[\lambda_{A}t = 2.303log\frac{{\text{ }N}_{A0}}{{\text{ }N}_{At}}\]

Similarly for isotope B

\[\lambda_{B}t = 2.303log\frac{{\text{ }N}_{B0}}{{\text{ }N}_{Bt}}\]

On subtracting

\[\begin{matrix} & \left( \lambda_{A} - \lambda_{B} \right)t = 2.303log\frac{{\text{ }N}_{A0}/N_{B0}}{{\text{ }N}_{At}/N_{Bt}} \\ \Rightarrow \ & \left( \lambda_{A} - \lambda_{B} \right)t = 2.303log\frac{\text{~}\text{initial ratio}\text{~}}{\text{~}\text{final ratio}\text{~}} \\ \Rightarrow \ & \left( \frac{1}{4 \times 10^{9}} - \frac{1}{2 \times 10^{9}} \right)t = 2.303log\frac{1}{100/1.02} \\ \Rightarrow \ & \text{~}\text{age}\text{~} = t = 1.83 \times 10^{10}\text{~}\text{years}\text{~} \end{matrix}\]

Q. 19 A given sample contains two types of atoms \(A\) and \(B\) in the ratio 3: 1. Atoms of type \(A\) undergo \(\alpha\)-decay with a half life of 30 days to form ' B ' while 'atoms of type B' undergo \(\alpha\)-decay with a half life of 45 days to from ' \(C\) ', which is stable. Calculate the time after which the activities of \(A\) and that of \(B\) are in the ratio 9:22

\[A\overset{\phantom{{\text{ }T}_{10} = 30\text{~}\text{days}\text{~}\ }}{\rightarrow}B\overset{\phantom{{\text{ }T}_{10} = 45\text{~}\text{days}\text{~}\ }}{\rightarrow}C\]

Sol. The radiactive decay series is given

Initially \(N_{A}(0):N_{B}(0) \smallsetminus 3:1\)

\[\begin{matrix} & \frac{dN_{A}}{dt} + \lambda_{A}N_{A} = 0 \\ & \frac{dN_{B}}{dt} = \lambda_{A}N_{A} - \lambda_{B}N_{B} \\ & \frac{dN_{C}}{dt} = \lambda_{B}N_{B} \\ & N_{A} = N_{A}(0)e^{- \lambda_{A}t} \\ & N_{B} = c_{1}e^{- \lambda_{B}t} + \frac{\lambda_{A}N_{A}(0)e^{- \lambda_{A}t}}{- \lambda_{A} + \lambda_{B}} \end{matrix}\]

Then we get, \(c_{1} = \frac{5}{2}{\text{ }N}_{0}\)
\[\therefore\ N_{A}(t) = \frac{3}{4}{\text{ }N}_{0}\left( \frac{1}{2} \right)^{\frac{1}{\tau_{1/2}}} = \frac{3}{4}{\text{ }N}_{0}\left( \frac{1}{2} \right)^{\frac{1}{30\text{~}\text{days}\text{~}}} \]and \(\ N_{B}(t) = \left\lbrack \frac{5}{2}{\text{ }N}_{0}\left( \frac{1}{2} \right)^{\frac{1}{45\text{~}\text{days}\text{~}}} - \frac{9}{4}{\text{ }N}_{0}\left( \frac{1}{2} \right)^{\frac{1}{30\text{~}\text{days}\text{~}}} \right\rbrack\)
Now, \(\frac{\lambda_{A}N_{A}}{\lambda_{B}N_{B}} = \frac{9}{22}\) i.e. \(\frac{N_{A}}{N_{B}} = \frac{3}{11}\)
or, \(\ \left( \frac{1}{2} \right)^{- t/90} = 2\)
or, \(\ t = 90\) days