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#### Page No 393:

#### Question 1:

When a Coolidge tube is operated for some time it becomes hot. Where does the heat come from?

#### Answer:

A Coolidge tube apparatus consists of a filament and a target. The filament is heated to produce electrons that are accelerated by applying an electric field between the filament and the target. When these accelerated electrons enter the target, they collide with the target atoms. In the process, the electrons lose their kinetic energy. A part of this kinetic energy is utilised for emitting X-rays and the remaining energy is absorbed by the target. Inside the target, the kinetic energy of the electrons is converted into heat energy. This raises the temperature of the target and hence, it heats the Coolidge tube.

#### Page No 393:

#### Question 2:

In a Coolidge tube, electrons strike the target and stop inside it. Does the target get more and more negatively charged as time passes?

#### Answer:

An electron emitted from the filament undergoes a number of collisions inside the material and loses its kinetic energy before coming to rest. This energy is utilised to give out photons or eject electrons from the atoms of the target. These electrons move to the battery connected to the circuit. Thus, the target does not get more and more negatively charged as time passes.

#### Page No 393:

#### Question 3:

Can X-rays be used for photoelectric effect?

#### Answer:

Yes, X-rays can be use for photoelectric effect. Photoelectric effect is the emission of electrons from a metal surface when the frequency of radiation is greater than the threshold frequency of the metal. For photoelectric effect using X-rays, the energy of the incoming X-ray photon should be greater than the work-function of the metal used.

#### Page No 393:

#### Question 4:

Can X-rays be polarised?

#### Answer:

Only transverse waves can be polarised. Since an X-ray is a transverse wave, it can be polarised.

#### Page No 393:

#### Question 5:

X-ray and visible light travel at the same speed in vacuum. Do they travel at the same speed in glass?

#### Answer:

Speed of light in any material medium is inversely proportional to the refractive index of the medium. Since refractive index of glass for X-ray is less than that for visible light, an X-ray will travel at a faster speed than visible light in glass.

#### Page No 393:

#### Question 6:

Characteristic X-rays may be used to identify the element from which they are being emitted. Can continuous X-rays be used for this purpose?

#### Answer:

Characteristic X-rays are emitted due to the transitions of electrons among different shells. The wavelength of the X-rays emitted in these transitions have definite value for a particular element. But continuous X-rays are emitted due to the conversion of kinetic energy of an electron into photon, which varies from collision to collision and is independent of material. Hence, continuous X-rays provide no information about the element from which they are being emitted.

#### Page No 393:

#### Question 7:

Is it possible that in a Coolidge tube characteristic Lα X-rays are emitted but not Kα X-rays?

#### Answer:

K_{α} X-rays are emitted due to the transition of an electron from the L shell to the K shell and L_{α} X-rays due to the transition of an electron from the M shell to the L shell. If K_{α} X-rays are not emitted, then the L shell will not be vacant to take the electron from the M shell. Hence, L_{α}_{}X-rays will not be emitted. Therefore, it is not possible that in a Coolidge tube, characteristic L_{α} X-rays are emitted but not K_{α}_{}X-rays.

#### Page No 393:

#### Question 8:

Can Lα X-ray of one material have shorter wavelength than Kα X-ray of another?

#### Answer:

An L_{α} X-ray is emitted when an electron jumps from the M to the L shell, and a K_{α} X-ray is emitted when an electron jumps from the L to the K shell. Less energy is involved when an electron jumps from the M to the L shell than when it jumps from the L to the K shell. Also, wavelength of a photon is inversely related to its energy. Therefore, an L_{α} X-ray has higher wavelength than a K_{α} X-ray for the same material.

#### Page No 393:

#### Question 9:

Can a hydrogen atom emit characteristic X-rays?

#### Answer:

The difference of energy levels in a hydrogen atom is small. Hence, it is not able to emit characteristic X-rays.

#### Page No 393:

#### Question 10:

Why is exposure to X-rays injurious to health but not exposure to visible light, when both are electromagnetic waves?

#### Answer:

X-rays have more penetrating power compared to visible light. As a result, they can penetrate the human body and can also damage the cells of the body. Prolonged exposure to X-rays can lead to cancer or genetic defects.

#### Page No 393:

#### Question 1:

An X-ray beam can be deflected

(a) by an electric field

(b) by a magnetic field

(c) by an electric field as well as a magnetic field

(d) neither by an electric field nor a magnetic field

#### Answer:

(d) neither by an electric field nor a magnetic field

Since X-rays do not have any charged particles, they are not deflected by an electric field or a magnetic field.

#### Page No 393:

#### Question 2:

Consider a photon of a continuous X-ray coming from a Coolidge tube. Its energy comes from

(a) the kinetic energy of the striking electron

(b) the kinetic energy of the free electrons of the target

(c) the kinetic energy of the ions of the target

(d) an atomic transition in the target

#### Answer:

(a) the kinetic energy of the striking electron

In an X-ray tube, electrons are emitted by the filament when it is heated. An electric field generated by a DC battery between the filament and the target makes the electrons hit the target atoms with a very high speed. As a result, the electrons lose their kinetic energy to eject photons, which leads to a continuous emission of X-rays.

#### Page No 393:

#### Question 3:

The energy of a photon of a characteristic X-ray from a Coolidge tube comes from

(a) the kinetic energy of the striking electron

(b) the kinetic energy of the free electrons of the target

(c) the kinetic energy of the ions of the target

(d) an atomic transition in the target

#### Answer:

(d) an atomic transition in the target

In an X-ray tube, electrons are emitted by the filament. These electrons are made to strike the filament by applying an electric field between the filament and the target. As a result of it, the kinetic energy of the electrons is lost to the target atoms. This energy is utilised by the target atoms to knock out an electron from the innermost shell. Consequently, the electron makes a transition from the higher energy state to this vacant shell. Due to this transition, the difference of energy of the two states gives photon of characteristic X-ray.

#### Page No 393:

#### Question 4:

If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled, the cutoff wavelength

(a) will remain unchanged

(b) will be doubled

(c) will be halved

(d) will become four times the original

#### Answer:

(c) will be halved

Cut off wavelength is given by

${\lambda}_{\mathrm{min}}=\frac{hc}{eV}$,

where *h *= Planck's constant

* c* = speed of light

*e *= charge on an electron

*V* = potential difference applied to the tube

When potential difference (*V*) applied to the tube is doubled, cutoff wavelength $\left(\lambda {\text{'}}_{\mathrm{min}}\right)$ is given by

$\lambda {\text{'}}_{\mathrm{min}}=\frac{hc}{e\left(2V\right)}$

$\Rightarrow \lambda {\text{'}}_{\mathrm{min}}=\frac{{\lambda}_{min}}{2}$

Cuttoff wavelength does not depend on the separation between the filament and the target.

Thus, cutoff wavelength will be halved if the the potential difference applied to the tube is doubled.

#### Page No 394:

#### Question 5:

If the current in the circuit for heating the filament is increased, the cutoff wavelength

(a) will increase

(b) will decrease

(c) will remain unchanged

(d) will change

#### Answer:

(c) will remain unchanged

Cut off wavelength is given by

${\lambda}_{\mathrm{min}}=\frac{hc}{eV}$,

where *h *= Planck's constant

* c* = speed of light

*e *= charge on an electron

*V* = potential difference applied to the tube

Clearly, the cutoff wavelength does not depend on the current in the circuit and temperature of the filament.

#### Page No 394:

#### Question 6:

Moseley's Law for characteristic X-ray is √*v* = a(*Z* − *b*). Here,

(a) both *a* and *b* are independent of the material

(b) *a* is independent but *b* depends on the material

(c) *b* is independent but *a* depends on the material

(d) both *a* and *b* depend on the material

#### Answer:

(a) both *a* and *b* are independent of the material

Moseley's Law for characteristic X-ray is √*v* = a(*Z* − *b*), where, *a* and *b* are constants independent of the material used.

#### Page No 394:

#### Question 7:

Frequencies of Kα X-rays of different materials are measured. Which one of the graphs in the figure may represent the relation between the frequency *v* and the atomic number *Z *?

Figure

#### Answer:

Using Moseley's Law,

$\sqrt{v}=a\left(Z-b\right)$,

where *v = *frequency of K_{α} X-ray

*Z* = atomic number

$\therefore v={a}^{2}{\left(Z-b\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(Z-b\right)}^{2}=\frac{v}{{a}^{2}}$

This is the equation of a parabola with some intercept on the axis, representing atomic number (Z). Hence, curve *d *represent this relation correctly.

#### Page No 394:

#### Question 8:

The X-ray beam emerging from an X-ray tube

(a) is monochromatic

(b) has all wavelengths smaller than a certain maximum wavelength

(c) has all wavelengths greater than a certain minimum wavelength

(d) has all wavelengths lying between a minimum and a maximum wavelength

#### Answer:

(c) has all wavelengths greater than a certain minimum wavelength

The X-ray beam emerging from an X-ray tube consists of wavelengths greater than a certain minimum wavelength called cutoff wavelength.

#### Page No 394:

#### Question 9:

One of the following wavelengths is absent and the rest are present in the X-rays coming from a Coolidge tube. Which one is the absent wavelength?

(a) 25 pm

(b) 50 pm

(c) 75 pm

(d) 100 pm

#### Answer:

(a) 25 pm

Cutoff wavelength (minimum) is absent in the X-rays coming from a Coolidge tube. Hence, the 25 pm wavelength is absent.

#### Page No 394:

#### Question 10:

The figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for the tube *A* in which the potential difference between the target and the filament is *V _{A}* and the atomic number of the target material is

*Z*. These quantities are

_{A}*V*and

_{B}*Z*for the other tube. Then,

_{B}(a)

*V*>

_{A}*V*,

_{B}*Z*>

_{A}*Z*

_{B}(b)

*V*>

_{A}*V*,

_{B}*Z*<

_{A}*Z*

_{B}(c)

*V*<

_{A}*V*,

_{B}*Z*>

_{A}*Z*

_{B}(d)

*V*>

_{A}*V*,

_{B}*Z*<

_{A}*Z*

_{B}Figure

#### Answer:

(b) *V _{A}* >

*V*,

_{B}*Z*<

_{A}*Z*

_{B}It is clear from the figure that the X-ray of tube A has less cutoff wavelength than the X-ray of tube B.

$\therefore {\lambda}_{A}{\lambda}_{B}$

Using Moseley's Law,

${Z}_{A}<{Z}_{B}$

$\lambda \propto \frac{1}{V}$, where

*V*is the voltage applied in the X-ray tube.

$\therefore {V}_{A}{V}_{B}$

#### Page No 394:

#### Question 11:

50% of the X-rays coming from a Coolidge tube are able to pass through a 0.1 mm thick aluminium foil. If the potential difference between the target and the filament is increased, the fraction of the X-rays passing through the same foil will be

(a) 0%

(b) < 50%

(c) 50 %

(d) > 50%

#### Answer:

(d) > 50%

The penetrating power of X-rays varies directly with the accelerating potential of the electrons (*V*) or the energy of the X-rays. So, if the potential difference between the target and the filament is increased, the fraction of the X-rays passing through the foil will also increase.

#### Page No 394:

#### Question 12:

50% of the X-ray coming from a Coolidge tube is able to pass through a 0.1 mm thick aluminium foil. The potential difference between the target and the filament is increased. The thickness of the aluminium foil that will allow 50% of the X-ray to pass through will be

(a) zero

(b) < 0.1 mm

(c) 0.1 mm

d) > 0.1 mm

#### Answer:

(d) > 0.1 mm

As we increase the accelerating potential between the filament and the target, the penetrating power of the X-ray will increase. As a result, the fraction of the X-ray passing through the foil will increase. But it is given that the same fraction of the X-ray is passing. Therefore, the thickness of the aluminium foil should be increased to maintain the same fraction of the X-ray that will pass through the foil.

#### Page No 394:

#### Question 13:

X-ray from a Coolidge tube is incident on a thin aluminium foil. The intensity of the X-ray transmitted by the foil is found to be *I*_{0}. The heating current is increased to increase the temperature of the filament. The intensity of the X-ray transmitted by the foil will be

(a) zero

(b) < *I*_{0}

(c) *I*_{0}

(d) > *I*_{0}

#### Answer:

(d) > *I*_{0}_{.}

We know that the intensity of an X-ray is directly proportional to the current through the X-ray tube. If the filament current is increased to increase the temperature of the filament, more electrons will be emitted from the filament per unit time. As a result, current in the X-ray tube will increase and consequently, the intensity of the X-ray will also increase.

#### Page No 394:

#### Question 14:

Visible light passing through a circular hole forms a diffraction disc of radius 0.1 mm on a screen. If an X-ray is passed through the same setup, the radius of the diffraction disc will be

(a) zero

(b) < 0.1 mm

(c) 0.1 mm

(d) > 0.1 m

#### Answer:

(b) < 0.1 mm

Radius of the diffraction disc is directly proportional to the wavelength of the light used for the given hole. We know that the wavelength of an X-ray is less than the wavelength of visible light. If an X-ray is passed through the same setup, the radius of the diffraction disc will be less than 0.1 m.

#### Page No 394:

#### Question 1:

For harder X-rays,

(a) the wavelength is higher

(b) the intensity is higher

(c) the frequency is higher

(d) the photon energy is higher.

#### Answer:

(c) the frequency is higher

(d) the photon energy is higher

Harder X-rays are the X-rays having low wavelengths. Since the frequency varies inversely with the wavelength, hard X-rays have high frequency.

Energy of a photon $\left(E\right)$ is given by

$E=\frac{hc}{\lambda}$

Here,*h* = Planck's constant*c* = Speed of light

$\lambda $ = Wavelength of light.

Clearly, energy varies inversely with wavelength. Therefore, the energy of the photon will be higher for the hard X-ray.

#### Page No 394:

#### Question 2:

Cutoff wavelength of X-rays coming from a Coolidge tube depends on the

(a) target material

(b) accelerating voltage

(c) separation between the target and the filament

(d) temperature of the filament.

#### Answer:

(b) accelerating voltage

Cutoff wavelength $\left({\lambda}_{\mathrm{min}}\right)$ is given by

${\lambda}_{\mathrm{min}}=\frac{hc}{eV}$

Here,*h* = Planck's constant*c* = Speed of light*V* = Accelerating voltage*e* = Charge of electron

Clearly, a cutoff wavelength depends on accelerating voltage. It does not depend on the target material, separation between the target and the temperature of the filament.

#### Page No 394:

#### Question 3:

Mark the correct options.

(a) An atom with a vacancy has smaller energy that a neutral atom.

(b) K X-ray is emitted when a hole makes a jump from the K shell to some other shell.

(c) The wavelength of K X-ray is smaller than the wavelength of L X-ray of the same material.

(d) The wavelength of K_{α} X-ray is smaller than the wavelength of K_{β} X-ray of the same material.

#### Answer:

(b) K X-ray is emitted when a hole makes a jump from the K shell to some other shell.

(c) The wavelength of K X-ray is smaller than the wavelength of L X-ray of the same material.

Energy of a vacant atom is higher than that of a neutral atom.

Hence, option (a) is incorrect.

K X-ray is emitted when an electron makes a jump to the K shell from some other shell. As a result, a positive charge hole is created in the outer shell. As the electron continuously moves to the K shell, the hole moves from the K shell to some other shell. Hence, option (b) is correct.

K X-ray is emitted due to the transition of an electron from the L or M shell to the* *K shell and L X-ray is emitted due to the transition of an electron from the M or N shell to the L shell. The energy involved in the transition from the L or M shell to the* *K shell is higher than the energy involved in the transition from the M or N shell to the L shell. Since the energy is inversely proportional to the wavelength, the wavelength of the K X-ray is smaller than the wavelength of the L X-ray of the same material. Hence, option (c) is correct.

If *E*_{K}, *E*_{L} and *E*_{M} are the energies of* *K*, *L and* *M* *shells, respectively, then the wavelength of K_{α} X-ray $\left({\lambda}_{1}\right)$ is given by

${\lambda}_{1}=\frac{hc}{{E}_{K}-{E}_{L}}$

Here,*h* = Planck's constant*c* = Speed of light

Wavelength of the K_{β} X-ray $\left({\lambda}_{2}\right)$ is given by

${\lambda}_{2}=\frac{hc}{{E}_{K}-{E}_{M}}$

As the difference of energies (*E*_{K}$-$* E*_{M}) is more than (*E _{K}* $-$

*E*), ${\lambda}_{2}$ is less than ${\lambda}_{1}$. Hence, option (d) is not correct.

_{L}#### Page No 394:

#### Question 4:

For a given material, the energy and wavelength of characteristic X-rays satisfy

(a) *E*(K_{α}) > *E*(K_{β}) > *E*(K_{γ})

(b) *E*(M_{α}) > *E*(L_{α}) > *E*(K_{α})

(c) λ(K_{α}) > λ(K_{β}) > λ(K_{γ})

(d) λ(M_{α}) > λ(L_{α}) > λ(K_{α}).

#### Answer:

(c) *λ*(K_{α}) > *λ*(K_{β}) > *λ*(K_{γ})

(d) *λ*(M_{α}) > *λ*(L_{α}) > *λ*(K_{α})

The K_{γ}_{}transition (from the N shell to the K shell) involves more energy than the K_{β} transition (from the M shell to* *the K shell), which has more energy than the K_{α} transition (from the L shell to* *the K shell).

As the energy varies inversely with the wavelength,*λ*(K_{α}) > *λ*(K_{β}) > *λ*(K_{γ})

The M_{α}_{}transition is due to the jumping of an electron from the* *N shell to the M shell and involves less energy than the L_{α} transition (from the M shell to the L shell), which involves less energy than the K_{α}_{}transition (from the L shell to the K* *shell).

As the energy varies inversely with the wavelength,*λ*(M_{α}) > *λ*(L_{α}) > *λ*(K_{α})

#### Page No 394:

#### Question 5:

The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation,

(a) the intensity increases

(b) the minimum wavelength increases

(c) the intensity remains unchanged

(d) the minimum wavelength decreases.

#### Answer:

(c) the intensity remains unchanged

(d) the minimum wavelength decreases

Cutoff wavelength $\left({\lambda}_{\mathrm{min}}\right)$ is given by

${\lambda}_{\mathrm{min}}=\frac{h\mathrm{c}}{eV}$

Here,*h* = Planck's constant

c = Speed of light*V* = Accelerating voltage*e* = Charge of electron

If the potential difference applied to an X-ray tube $\left(V\right)$ is increased, then the minimum wavelength (cutoff wavelength) gets decreased.

Intensity is not affected by the potential difference applied.

#### Page No 394:

#### Question 6:

When an electron strikes the target in a Coolidge tube, its entire kinetic energy

(a) is converted into a photon

(b) may be converted into a photon

(c) is converted into heat

(d) may be converted into heat.

#### Answer:

(b) may be converted into a photon

(d) may be converted into heat

When an electron strikes the target in a Coolidge tube, the kinetic energy of the electron is used in two ways. Some part of the kinetic energy is converted into a photon, while the remaining part gets converted into heat when the electron makes collisions with the atoms of the target. However, the amount of kinetic energy appearing as the photon vary from collision to collision.

#### Page No 395:

#### Question 7:

X-ray incident on a material

(a) exerts a force on it

(b) transfers energy to it

(c) transfers momentum to it

(d) transfers impulse to it.

#### Answer:

(a) exerts a force on it

(b) transfers energy to it

(c) transfers momentum to it

(d) transfers impulse to it.

An X-ray exerts force on the material on which it is incident. For example, it can penetrate into metals. An X-ray also transfers energy, momentum and impulse to the material on which it incident. Therefore, all options are correct.

#### Page No 395:

#### Question 8:

Consider a photon of continuous X-ray and a photon of characteristic X-ray of the same wavelength. Which of the following is/are different for the two photons?

(a) Frequency

(b) Energy

(c) Penetrating power

(d) Method of creation

#### Answer:

(d) Method of creation

Let ${\lambda}_{1}$ and ${\lambda}_{2}$ be the wavelengths of the photons of continuous and characteristic X-rays, respectively.

Given:

${\lambda}_{1}={\lambda}_{2}=\lambda $

Now, frequency $\left(\nu \right)$ is given by

$v=\frac{c}{{\lambda}_{1}}=\frac{c}{{\lambda}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =\frac{c}{\lambda}$

Hence, the frequency of both the photons is the same.

Energy of a photon $\left(E\right)$ is given by

$E=hv$

As the frequency of both the photons is the same, they will have the same energy.

The photon of the continuous X-ray and the photon of the characteristic X-ray consist of the same penetrating power.

The photon of the characteristic X-ray is created because of the transition of an electron from one shell to another. The photon of the continuous X-ray is created because of the conversion of the kinetic energy of an electron into a photon of electromagnetic radiation.

#### Page No 395:

#### Question 1:

Find the energy, the frequency and the momentum of an X-ray photon of wavelength 0.10 nm.

#### Answer:

Given:

Wavelength of the X-ray photon, *λ *= 0.1 nm

Speed of light, *c* = 3$\times $10^{8} m/s

(a) Energy of the photon $\left(E\right)$ is given by

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1242\mathrm{eV}}{0.1}\phantom{\rule{0ex}{0ex}}\Rightarrow E=12420\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow E=12.42\mathrm{keV}\approx 12.4\mathrm{keV}$

(b) Frequency is given by

$\nu =\frac{c}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =\frac{3\times {10}^{8}}{0.1\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =3\times {10}^{18}\mathrm{Hz}$

(c) Momentum of the photon $\left(P\right)$ is given by

$P=\frac{E}{c}\phantom{\rule{0ex}{0ex}}\Rightarrow P=\frac{12.4\times {10}^{3}\times 1.6\times {10}^{-19}}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=6.613\times {10}^{-24}\mathrm{kg}-\mathrm{m}/\mathrm{s}\approx 6.62\times {10}^{-24}\mathrm{g}-\mathrm{m}/\mathrm{s}$

#### Page No 395:

#### Question 2:

Iron emits K_{α} X-ray of energy 6.4 keV. Calculate the times taken by an iron K_{α} photon to cross through a distance of 3 km.

#### Answer:

Given:

Energy of the X-ray, *E* = 6.4 keV

Distance travelled by the photon, *d* = 3 km = 3 $\times $ 10^{3} m

Time taken by the photon to cross the distance of 3 km is given by

$t=\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{3\times {10}^{3}}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}t={10}^{-5}\mathrm{s}=10\times {10}^{-6}\mathrm{s}\phantom{\rule{0ex}{0ex}}t=10\mathrm{\mu s}$

Both the K_{α} photon and X-ray will take the same time, that is, 10 $\mathrm{\mu s}$, to cross the distance of 3 km.

#### Page No 395:

#### Question 3:

Find the cutoff wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV.

#### Answer:

Given:

Voltage of the X-ray tube, *V = *30 kV

Cutoff wavelength for continuous X-rays $\left(\lambda \right)$ is given by

$\lambda =\frac{hc}{e\mathrm{V}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}h=\mathrm{Planck}\text{'}s\mathrm{constant}\phantom{\rule{0ex}{0ex}}V=\mathrm{Voltage}\mathrm{of}\mathrm{the}\mathrm{X}-\mathrm{ray}\mathrm{tube}\phantom{\rule{0ex}{0ex}}c=\mathrm{Speed}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}\therefore \lambda =\frac{1242\mathrm{eV}-\mathrm{nm}}{\mathrm{e}\times 30\times {10}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =414\times {10}^{-4}\mathrm{nm}=41.4\mathrm{pm}$

#### Page No 395:

#### Question 4:

What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?

#### Answer:

Given:

Wavelength of the X-ray, $\lambda $ = 0.10 nm

Planck's constant,* h* = 6.63$\times $10^{$-34$} J-s

Speed of light, *c* = 3$\times {10}^{8}$ m/s

Minimum wavelength is given by

${\lambda}_{\mathrm{min}}=\frac{hc}{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{hc}{e{\lambda}_{\mathrm{min}}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{1.6\times {10}^{-19}\times {10}^{-10}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=12.43\times {10}^{3}\mathrm{V}=12.4\mathrm{kV}$

Maximum energy of the photon $\left(E\right)$ is given by*E *= $\frac{hc}{\lambda}$

$\Rightarrow E=\frac{6.68\times {10}^{-34}\times 3\times {10}^{8}}{{10}^{-10}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=19.89\times {10}^{-16}\phantom{\rule{0ex}{0ex}}\Rightarrow E=1.989\times {10}^{-15}\approx 2\times {10}^{-15}\mathrm{J}$

#### Page No 395:

#### Question 5:

The X-ray coming from a Coolidge tube has a cutoff wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target.

#### Answer:

Given:

Cutoff wavelength of the Coolidge tube, $\lambda $ = 80 pm

Energy of the electron hitting the target $\left(E\right)$ is given by

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}$

Here,*h *= Planck's constant*c* = Speed of light

$\lambda $ = Wavelength of light

$\therefore E=\frac{1242\mathrm{eV}-\mathrm{nm}}{80\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1242\times {10}^{-9}\mathrm{eV}}{80\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=15.525\times {10}^{3}\mathrm{eV}\approx 15.5\mathrm{keV}$

#### Page No 395:

#### Question 6:

If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cutoff wavelength decrease?

#### Answer:

Let $\lambda $ be the cut off wavelength and *V* be the operating potential in the X-ray tube.

Then,

$\lambda =\frac{hc}{V}\phantom{\rule{0ex}{0ex}}$

Here,*h* = Planck's constant*c *= Speed of light

If the operating voltage is increased by 1%, then the new operating voltage (*V*') will be given by

$V\text{'}=\mathrm{V}+\frac{1}{100}\times \mathrm{V}\phantom{\rule{0ex}{0ex}}V\text{'}=1.01\mathrm{V}$

Cut off wavelength $\left(\lambda \text{'}\right)$ on increasing the operating voltage is given by

$\lambda \text{'}=\frac{hc}{1.01V}=\frac{\lambda}{1.01}\phantom{\rule{0ex}{0ex}}\therefore \lambda \text{'}-\lambda =\frac{0.01}{1.01}\lambda $

Percentage change in the wavelength is given by

$\frac{0.01\lambda}{1.01\times \lambda}\times 100=\frac{1}{1.01}\phantom{\rule{0ex}{0ex}}=0.9901$

= 1 (approx.)

#### Page No 395:

#### Question 7:

The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.

#### Answer:

Given:

Distance between the filament and the target in the X-ray tube, *d* = 1.5 m

Cut off wavelength, $\lambda $ = 30 pm

Energy $\left(E\right)$ is given by

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}$

Here,*h *= Planck's constant*c* = Speed of light

$\lambda $ = Wavelength of light

Thus, we have

$E=\frac{1242\mathrm{eV}-\mathrm{nm}}{30\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1242\times {10}^{-9}}{30\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=41.4\times {10}^{3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Electric}\mathrm{field}=\frac{\mathrm{V}}{d}=\frac{41.4\times {10}^{3}}{1.5}\phantom{\rule{0ex}{0ex}}=27.6\times {10}^{3}\mathrm{V}/\mathrm{m}\phantom{\rule{0ex}{0ex}}=27.6\mathrm{kV}/\mathrm{m}$

#### Page No 395:

#### Question 8:

The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

#### Answer:

Let $\lambda $ be the initial wavelength, *V* be the initial potential, $\lambda \text{'}$ be the new wavelength and *V*' be the new operating voltage when the operating voltage is increased in the X-ray tube.

Given:

$\lambda \text{'}$ = $\lambda $$-$26 pm*V*' = 1.5 V

Energy $\left(E\right)$ is given by

$E=\frac{hc}{\lambda}$

$\Rightarrow eV=\frac{hc}{\lambda}$

Here,*h* = Planck's constant*c* = Speed of light

$\lambda $ = Wavelength of light*V* = Operating potential

$\therefore \lambda =\frac{hc}{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda V=\lambda \text{'}\mathrm{V}\text{'}\left[\because \mathrm{\lambda}\propto \frac{1}{V}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda V=\left(\lambda -26\right)\times 1.5\mathrm{V}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.5\lambda =26\times 1.5\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =26\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =78\mathrm{pm}$

Hence, the initial wavelength is 78$\times $10^{$-$12} m.

Now, the operating voltage (*V*) is given by

$V=\frac{hc}{e\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{1.6\times {10}^{-19}\times 78\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=0.15937\times {10}^{5}\phantom{\rule{0ex}{0ex}}\Rightarrow V=15.9\mathrm{kV}$

#### Page No 395:

#### Question 9:

The electron beam in a colour TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?

#### Answer:

Given:

Potential of the electron beam, *V *= 32 kV = 32 × 10^{3} V

Energy, $E=32\times {10}^{3}\mathrm{eV}$

Wavelength of the X-ray photon $\left(\lambda \right)$ is given by

$\lambda =\frac{h\mathrm{c}}{E}\phantom{\rule{0ex}{0ex}}$

Here,*h* = Planck's constant

c = Speed of light

$\therefore \lambda =\frac{hc}{E}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{1242\mathrm{eVnm}}{32\times {10}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =38.8\times {10}^{-3}\mathrm{nm}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =38.8\mathrm{pm}$

#### Page No 395:

#### Question 10:

When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of 9.7 × 10^{18} Hz. Calculate the value of Planck constant from these data.

#### Answer:

Given:

Potential applied to the X-ray tube, *V* = 40 kV

Frequency of the X-ray, *v = *9.7 × 10^{18} Hz

Wavelength $\left(\lambda \right)$ is given by

$\lambda =\frac{hc}{eV}\phantom{\rule{0ex}{0ex}}$

Here,*h* = Planck's constant*c* = Speed of light*e* = 1.6 $\times $ 10^{$-19$}

$\therefore \lambda =\frac{hc}{e\mathrm{V}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\lambda}{c}=\frac{h}{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{h}{eV}\left(\because v=\frac{c}{\lambda}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{eV}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{40\times {10}^{3}}{9.7\times {10}^{18}}\times e\phantom{\rule{0ex}{0ex}}\Rightarrow h=4.12\times {10}^{-15}e\mathrm{Vs}$

#### Page No 395:

#### Question 11:

An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest there wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

#### Answer:

Given:

Potential of the X-ray tube, *V* = 40 kV = 40 × 10^{3} V

Energy* *= 40 × 10^{3} eV

Energy utilised by the electron is given by*E *=$\frac{70}{100}\times 40\times {10}^{3}$ = 28 $\times $ 10^{3} eV

Wavelength $\left(\lambda \right)$ is given by

$\lambda =\frac{hc}{E}$

Here,*h* = Planck's constant*c *= Speed of light*E* = Energy of the electron

$\therefore \lambda =\frac{hc}{E}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{1242\mathrm{eV}-\mathrm{nm}}{28\times {10}^{3}\mathrm{eV}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{1242\times {10}^{-9}\mathrm{eV}}{28\times {10}^{3}\mathrm{eV}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =44.35\times {10}^{-12}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =44.35\mathrm{pm}$

For the second wavelength,*E* = 70% (Leftover energy)

$=\frac{70}{100}\times (40-28){10}^{3}\phantom{\rule{0ex}{0ex}}=\frac{70}{100}\times 12\times {10}^{3}\phantom{\rule{0ex}{0ex}}=84\times {10}^{2}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}\lambda =\frac{hc}{E}=\frac{1242}{8.4\times {10}^{3}}\phantom{\rule{0ex}{0ex}}=147.86\times {10}^{-3}\mathrm{nm}\phantom{\rule{0ex}{0ex}}=147.86\mathrm{pm}=148\mathrm{pm}$

For the third wavelength,

$E=\frac{70}{100}(12-8.4)\times {10}^{3}\phantom{\rule{0ex}{0ex}}=7\times 3.6\times {10}^{2}=25.2\times {10}^{2}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}\lambda =\frac{hc}{E}=\frac{1242}{25.2\times {10}^{2}}\phantom{\rule{0ex}{0ex}}=49.2857\times {10}^{-2}\phantom{\rule{0ex}{0ex}}=493\mathrm{pm}$

#### Page No 395:

#### Question 13:

The K_{β} 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.

#### Answer:

Given:

Wavelength of K_{β} X-ray of argon, *λ *= 0.36 nm

Energy needed to ionise an argon atom = 16 eV

Energy of K_{β} X-ray of argon $\left(E\right)$ is given by

$E=\frac{1242}{0.36}=3450\mathrm{eV}$

Energy needed to knock out an electron from K shell*E*_{K} = (3450 + 16) eV*E*_{K} = 3466 eV $\approx $ 3.47 keV

#### Page No 395:

#### Question 14:

The K_{α} X-rays of aluminium (*Z* = 13) and zinc (*Z* = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley's law √v =* a*(*Z *− *b*) to find the wavelengths of the K_{α} X-ray of iron (Z = 26).

#### Answer:

Given:

Wavelength of K_{α} X-rays of aluminium, *λ*_{1} = 887 pm

Frequency of X-rays of aluminium is given by

${\nu}_{a}=\frac{c}{\lambda}\phantom{\rule{0ex}{0ex}}{\nu}_{a}=\frac{3\times {10}^{8}}{887\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}{\nu}_{a}=3.382\times {10}^{17}\phantom{\rule{0ex}{0ex}}{\nu}_{a}=33.82\times {10}^{16}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}$

Wavelength of K_{α} X-rays of zinc, ${\lambda}_{2}$ = 146 pm

Frequency of X-rays of zinc is given by

${\nu}_{z}=\frac{3\times {10}^{8}}{146\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}{\nu}_{z}=0.02055\times {10}^{20}\phantom{\rule{0ex}{0ex}}{\nu}_{z}=2.055\times {10}^{18}\mathrm{Hz}$

We know

*$\sqrt{\nu}$* = *a*(*Z* − *b*)

For aluminium,

5.815 × 10^{8} = *a*(13 − *b*) ...(1)

For zinc,

1.4331 × 10^{9} = *a*(30 − *b*) ...(2)

Dividing (1) by (2)

$\frac{13-b}{30-b}=\frac{5.815\times {10}^{-1}}{1.4331}\phantom{\rule{0ex}{0ex}}=0.4057\phantom{\rule{0ex}{0ex}}\Rightarrow 30\times 0.4057-0.4057b=13-b\phantom{\rule{0ex}{0ex}}\Rightarrow 12.171-0.4057b+b=13\phantom{\rule{0ex}{0ex}}b=\frac{0.829}{0.5943}=1.39491\phantom{\rule{0ex}{0ex}}a=\frac{5.815\times {10}^{8}}{11.33}\phantom{\rule{0ex}{0ex}}=0.51323\times {10}^{8}=5\times {10}^{7}$

For Fe,

Frequency $\left(\nu \text{'}\right)$ is given by*$\nu $*' = 5× 10^{7} (26 − 1.39)

= 5 × 24.61 × 10^{7}

= 123.05 × 10^{7}*$\nu $*' = $\frac{c}{\lambda}$

Here,* c* = speed of light

$\lambda $ = Wavelength of light

$\therefore \frac{c}{\lambda}=5141.3\times {10}^{14}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{3\times {10}^{8}}{5141.3\times {10}^{14}}\phantom{\rule{0ex}{0ex}}=0.000198\times {10}^{-5}\mathrm{m}\phantom{\rule{0ex}{0ex}}=198\times {10}^{-12}=198\mathrm{pm}$

#### Page No 395:

#### Question 15:

A certain element emits K_{α} X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.

#### Answer:

Given:

Energy of K_{α} X-ray, *E* = 3.69 keV = 3690 eV

Wavelength $\left(\lambda \right)$ is given by

$\lambda =\frac{hc}{E}\phantom{\rule{0ex}{0ex}}\lambda =\frac{1242}{3690}\phantom{\rule{0ex}{0ex}}\lambda =0.33658\mathrm{nm}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =0.34\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

From Moseley's equation,

$\sqrt{\frac{c}{\lambda}}=a(Z-b)$

Here, *c* = speed of light

$\lambda $ = wavelength of light

*Z* = atomic number of element

On substituting the respective values,

$\sqrt{\frac{3\times {10}^{8}}{0.34\times {10}^{-9}}}=5\times {10}^{7}(Z-1.39)\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{8.82\times {10}^{17}}=5\times {10}^{7}(Z-1.39)\phantom{\rule{0ex}{0ex}}\Rightarrow 9.39\times {10}^{8}=5\times {10}^{7}(Z-1.39)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{93.9}{5}=Z-1.39\phantom{\rule{0ex}{0ex}}\Rightarrow Z=\frac{93.9}{5}+1.39\phantom{\rule{0ex}{0ex}}=20.17=20$

So, the element is calcium.

#### Page No 395:

#### Question 16:

The K_{β} X-rays from certain elements are given below. Draw a Moseley-type plot of √v versus Z for K_{β} radiation.

Element | Ne | P | Ca | Mn | Zn | Br |

Energy (keV) | 0.858 | 2.14 | 4.02 | 6.51 | 9.57 | 13.3 |

#### Answer:

We can directly get value of *v *by energy frequency relation.*hv* = energy

Here, *h* = Planck's constant

$v=\frac{\mathrm{Energy}(\mathrm{in}\mathrm{keV})}{h}$

The required graph is as follows:

#### Page No 395:

#### Question 17:

The K_{α} and K_{β} X-rays of molybdenum have wavelengths 0.71 A and 0.63 A respectively. Find the wavelength of L_{α} X-ray of molybdenum.

#### Answer:

Given:

Wavelength of K_{α} X-rays, $\lambda $_{1} = 0.71 A

Wavelength of K_{β} X-rays, ${\lambda}_{2}$ = 0.63 A

For L_{a} X-ray of molybdenum,

*a* = 57

*b* = 1

From Moseley's equation,

$\sqrt{v}=a(\mathrm{Z}-b)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Here,* v* = frequency of X-ray

*Z *= atomic number of the element

$\sqrt{v}=a(\mathrm{Z}-b)\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{v}=a(57-1)\phantom{\rule{0ex}{0ex}}=a\times 56...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{C}u\left(29\right),\phantom{\rule{0ex}{0ex}}\sqrt{1.88\times {10}^{18}}=a(29-1)=28a...\left(2\right)\phantom{\rule{0ex}{0ex}}$

Dividing (1) and (2)

$\sqrt{\frac{v}{1.88\times {10}^{18}}}=\frac{a\times 56}{a\times 28}=2\phantom{\rule{0ex}{0ex}}\Rightarrow v=1.88\times {10}^{18}\times (2{)}^{2}\phantom{\rule{0ex}{0ex}}=4\times 1.88\times {10}^{18}\phantom{\rule{0ex}{0ex}}=7.52\times {10}^{18}\mathrm{Hz}$

#### Page No 395:

#### Question 18:

The K_{α} and K_{β} X-rays of molybdenum have wavelengths 0.71 A and 0.63 A respectively. Find the wavelength of L_{α} X-ray of molybdenum.

#### Answer:

Given:

Wavelength of K_{α}_{}X-rays of molybdenum, ${\lambda}_{a}$ = 0.71 A

Wavelength of K_{β} X-rays of molybdenum, ${\lambda}_{b}$ = 0.63 A

Energy of K_{α}X-rays$\left({K}_{a}\right)$ is given by*K*_{a} = *E*_{K}$-$*E*_{L} .....(1)

Energy of K_{β} X-rays $\left({K}_{\beta}\right)$ is given by

${K}_{\beta}={E}_{K}-{E}_{M}$ ....(2)

Energy of L_{a} X-ray $\left({L}_{a}\right)$ is given by

${K}_{L}={E}_{L}-{E}_{M}$

Subtracting (2) from (1),

${K}_{\mathrm{\alpha}}-{K}_{\mathrm{\beta}}={E}_{\mathrm{M}}-{E}_{\mathrm{L}}=-{K}_{L}\phantom{\rule{0ex}{0ex}}\mathrm{or}{K}_{L}={K}_{\mathrm{\beta}}-{K}_{\mathrm{\alpha}}\phantom{\rule{0ex}{0ex}}{K}_{L}=\frac{3\times {10}^{8}}{0.63\times {10}^{-10}}-\frac{3\times {10}^{8}}{0.71\times {10}^{-10}}\phantom{\rule{0ex}{0ex}}{K}_{L}=4.761\times {10}^{-18}-4.225\times {10}^{18}\phantom{\rule{0ex}{0ex}}{K}_{L}=0.536\times {10}^{18}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}\mathrm{Again},\mathrm{\lambda}=\frac{3\times {10}^{8}}{0.536\times {10}^{-18}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=5.6\times {10}^{-10}=5.6\mathrm{\AA}$

#### Page No 395:

#### Question 19:

The wavelengths of K_{α} and L_{α} X-rays of a material are 21.3 pm and 141 pm respectively. Find the wavelength of K_{β} X-ray of the material.

#### Answer:

Given:

Wavelength of K_{α} X-ray, ${\lambda}_{1}$ = 21.3 pm

Wavelength of L_{α} X-ray, ${\lambda}_{2}$ = 141 pm

Energy of K_{α} X-ray $\left({E}_{1}\right)$ is given by

${E}_{1}=\frac{1242}{21.3\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=58.309\times {10}^{3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

Energy of L_{α} X-ray $\left({E}_{2}\right)$ is given by

${E}_{2}=\frac{1242}{141\times {10}^{-5}}\phantom{\rule{0ex}{0ex}}=8.8085\times {10}^{3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

Energy of K_{β} X-ray $\left({E}_{3}\right)$ will be

${E}_{3}={E}_{1}+{E}_{2}\phantom{\rule{0ex}{0ex}}{E}_{3}=(58.309+8.809)\times {10}^{3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}{E}_{3}=67.118\times {10}^{3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

Wavelength of K_{β} X-ray $\left(\lambda \right)$ is given by

$\mathrm{\lambda}=\frac{hc}{{E}_{3}}=\frac{1242}{67.118\times {10}^{3}}\phantom{\rule{0ex}{0ex}}=18.5\times {10}^{-3}\mathrm{nm}=18.5\mathrm{pm}$

#### Page No 395:

#### Question 20:

The energy of a silver atom with a vacancy in K shell is 25.31 keV, in L shell is 3.56 keV and in M shell is 0.530 keV higher than the energy of the atom with no vacancy. Find the frequency of K_{α}, K_{β} and L_{α} X-rays of silver.

#### Answer:

Given:

Energy of electron in the K shell, *E*_{k} = 25.31 keV

Energy of electron in the L shell, *E*_{L} = 3.56 keV

Energy of electron in the M shell, *E*_{M} = 0.530 keV

Let *f* be the frequency of K$\alpha $ X-ray and *f*_{0} be the frequency of K_{β} X-ray.

Let *f*_{1} be the frequency of L_{α} X-rays of silver.

∴ K_{α} = *E*_{K} − *E*_{L} = *hf*

Here*, h* = Planck constant

*f* = frequency of K$\alpha $ X-ray

$f=\frac{{E}_{\mathrm{K}}-{E}_{\mathrm{L}}}{h}\phantom{\rule{0ex}{0ex}}f=\frac{(25.31-3.56)}{6.63\times {10}^{-34}}\times 1.6\times {10}^{-19}\times {10}^{3}\phantom{\rule{0ex}{0ex}}f=\frac{21.75\times {10}^{3}\times {10}^{15}}{6.67}\phantom{\rule{0ex}{0ex}}f=5.25\times {10}^{18}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{\beta}}={E}_{\mathrm{K}}-{E}_{\mathrm{M}}=h{f}_{0}\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{0}=\frac{{E}_{\mathrm{K}}-{E}_{\mathrm{M}}}{h}\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{0}=\frac{(25.31-0.53)}{6.67\times {10}^{-34}}\times {10}^{3}\times 1.6\times {10}^{-19}\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{0}=5.985\times {10}^{18}\mathrm{Hz}$

${K}_{L}={E}_{L}-{E}_{M}=h{f}_{1}\phantom{\rule{0ex}{0ex}}{f}_{1}=\frac{{E}_{L}-{E}_{M}}{h}\phantom{\rule{0ex}{0ex}}{f}_{1}=\frac{3.56-0.530}{6.63\times {10}^{-34}}\times {10}^{3}\times 1.6\times {10}^{-19}\phantom{\rule{0ex}{0ex}}{f}_{1}=7.32\times {10}^{17}\mathrm{Hz}$

#### Page No 395:

#### Question 21:

Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows.

Cell containing vacancy | K | L | M |

Energy in keV | 69.5 | 11.3 | 2.3 |

#### Answer:

Let the potential required that may be applied across the X-ray tube without emitting any characteristic K or L X-ray be *V*.

∴ Energy of electron = *eV*

This amount of energy is equal to the energy of L shell.

So, the maximum potential difference that can be applied without emitting any electron is 11.3 kV.

#### Page No 395:

#### Question 22:

The electric current in an X-ray tube (from the target to the filament) operating at 40 kV is 10 mA. Assume that on an average, 1% of the total kinetic energy of the electron hitting hte target are converted into X-rays.

(a) What is the total power emitted as X-rays and (b) how much heat is produced in the target every second?

#### Answer:

Given:

V = 30 KV*i* = 10mA

1% of T_{KE} (total kinetic energy) = X-ray*i* = *ne$\mathrm{or}n=\frac{{10}^{-2}}{1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=0.625\times {10}^{17}(\mathrm{number}\mathrm{of}\mathrm{electrons})$*

KE of one electron = eV

$=1.6\times {10}^{-19}\times 40\times {10}^{3}\phantom{\rule{0ex}{0ex}}=6.4\times {10}^{-15}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{\mathrm{KE}}=0.625\times 6.4\times {10}^{17}\times {10}^{-15}\phantom{\rule{0ex}{0ex}}=4\times {10}^{2}\mathrm{J}$

(b) Heat produced in target per second = 400 − 4 = 396 J

#### Page No 395:

#### Question 23:

Heat at the rate of 200 W is produced in an X-ray tube operating at 20 kV. Find the current in the circuit. Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays.

#### Answer:

Given:

Heat produced/second = Power (*P*) = 200 W

Potential in the X ray tube, *V* = 20 kV

We know

Power, *P* = *VI*

Here, V = potential difference

*I = *current

But *I* = $\frac{ne}{t}$

Where, *e* = charge on electron

* t *= time

*n* = no of electrons

$\therefore P=\frac{ne\mathrm{V}}{t}\phantom{\rule{0ex}{0ex}}\Rightarrow 200=\frac{neV}{t}\phantom{\rule{0ex}{0ex}}\Rightarrow (ne/t)\mathrm{V}=200\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{200}{V}\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{200}{20\times {10}^{3}}=10\mathrm{mA}$

#### Page No 395:

#### Question 12:

The wavelength of K_{α} X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of K_{α} X-ray?

#### Answer:

Given:

Wavelength of X-ray of tungsten, $\lambda $ = 21.3 pm

Energy required to take out electron from the L shell of a tungsten atom, *E*_{L} = 11.3 keV

Voltage required to take out electron from the L shell of a tungsten atom, *V*_{L} = 11.3 kV

Let *E*_{K} and *E*_{L} be the energies of K and L, respectively.

${E}_{\mathrm{K}}-{E}_{\mathrm{L}}=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{Here},h=\mathrm{Planck}\text{'}\mathrm{s}\mathrm{constant}\phantom{\rule{0ex}{0ex}}c=\mathrm{Speed}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{K}}-{E}_{L}=\frac{1242\mathrm{eV}-\mathrm{nm}}{21.3\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{K}}-{E}_{L}=\frac{1242\times {10}^{-9}\mathrm{eV}}{21.3\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{K}}-{E}_{L}=58.309\mathrm{keV}\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{L}}=11.3\mathrm{keV}\phantom{\rule{0ex}{0ex}}\therefore {E}_{\mathrm{K}}=69.609\mathrm{keV}$

Thus, the accelerating voltage across an X-ray tube that allows the production of K_{α} X-ray is given by*V*_{K} = 69.609 kV

#### Page No 396:

#### Question 24:

Continuous X-rays are made to strike a tissue paper soaked with polluted water. The incoming X-rays excite the atoms of the sample by knocking out the electrons from the inner shells. Characteristic X-rays are analysed and the intensity is plotted against the wavelength. Assuming that only K_{α} intensities are detected, list the elements present in the sample from the plot. Use Moseley's equation v − (25 × 10^{14}Hz)(Z − 1)^{2}.

Figure

#### Answer:

Given:*f *= (25 × 10^{14} Hz) (Z − 1)^{2
$\mathrm{or}\frac{c}{\lambda}=25\times {10}^{14}(\mathrm{Z}-1{)}^{2}\phantom{\rule{0ex}{0ex}}(\mathrm{a})\frac{3\times {10}^{8}}{78.9\times {10}^{-12}\times 25\times {10}^{14}}=(\mathrm{Z}-1{)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{or}(\mathrm{Z}-1{)}^{2}=.0.001520\times {10}^{6}=1520\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Z}-1=38.98\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}=39.98=40\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{it}\mathrm{is}(\mathrm{Zr}).\phantom{\rule{0ex}{0ex}}(\mathrm{b})\frac{3\times {10}^{8}}{146\times {10}^{-12}\times 25\times {10}^{14}}=(\mathrm{Z}-1{)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{or}(\mathrm{Z}-1{)}^{2}=0.00082219\times {10}^{6}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}-1=28.669\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}=29.669=30\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{it}\mathrm{is}(\mathrm{Zn}).\phantom{\rule{0ex}{0ex}}(\mathrm{c})\frac{3\times {10}^{8}}{158\times {10}^{-12}\times 25\times {10}^{4}}=(Z-1{)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{or}(\mathrm{Z}-1{)}^{2}=0.0007594\times {10}^{4}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}-1=27.5559\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}=28.5589=29\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{it}\mathrm{is}(\mathrm{Cu}).\phantom{\rule{0ex}{0ex}}(\mathrm{d})\frac{3\times {10}^{8}}{198\times {10}^{-12}\times 25\times {10}^{4}}=(\mathrm{Z}-1{)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{or}(\mathrm{Z}-1{)}^{2}=0.000606\times {10}^{4}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}-1=24.6162\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{Z}=25.6162=26\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{it}\mathrm{is}(\mathrm{Fe}).$}

#### Page No 396:

#### Question 25:

A free atom of iron emits K_{α} X-rays of energy 6.4 keV. Calculate the recoil kinetic energy of the atom. Mass of an iron atom = 9.3 × 10^{−26} kg.

#### Answer:

Here, energy of photon = E

E = 6.4 KeV = 6.4 × 10^{3} eV

Momentum of photon

$=\frac{\mathrm{E}}{c}=\frac{6.4\times {10}^{3}}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}=3.41\times {10}^{-24}\mathrm{kgm}/\mathrm{sec}$

According to collision theory,

Momentum of photon = Momentum of atom

∴ Momentum of atom, P = 3.41 × 10^{−24} kgm/sec

Recoil KE of atom = P^{2}/2M

$=\frac{(3.41\times {10}^{-24}{)}^{2}e\mathrm{V}}{\left(20\right)9.3\times {10}^{-26}\times 1.6\times {10}^{-19}}=3.9e\mathrm{V}$

[1 Joule = 1.6 × 10^{−19} eV]

#### Page No 396:

#### Question 26:

The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength (1/λ) of the light falling on the cathode. The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength (1/λ) of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find its value.

#### Answer:

V_{0} - Stopping Potential

K - Potential difference across X-ray tube

λ - Wavelength

λ - Cut difference Wavelength

$e{\mathrm{V}}_{0}=hf-h{f}_{0}\phantom{\rule{0ex}{0ex}}\lambda =\frac{hc}{e\mathrm{V}}\phantom{\rule{0ex}{0ex}}e{\mathrm{V}}_{0}=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{V\lambda}=\frac{hc}{e}\phantom{\rule{0ex}{0ex}}\mathrm{or}{\mathrm{V}}_{0}\mathrm{\lambda}=\frac{hc}{e}$

Here, the slopes are same.

i.e. V_{0}λ = Vλ

$\frac{hc}{e}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=1.242\times {10}^{-6}\mathrm{Vm}$

#### Page No 396:

#### Question 27:

Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through a Young's double slit and the interference pattern is observed on a photographic plate placed 40 cm away from the slit. What should be the separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm?

#### Answer:

Given:

$\lambda =10\mathrm{pm}=100\times {10}^{-12}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{D}=40\mathrm{cm}=40\times {10}^{-2}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{\beta}=0.1\mathrm{mm}=0.1\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{\beta}=\frac{\mathrm{\lambda D}}{d}\phantom{\rule{0ex}{0ex}}d=\frac{\mathrm{\lambda D}}{\mathrm{\beta}}\phantom{\rule{0ex}{0ex}}=\frac{100\times {10}^{-12}\times 40\times {10}^{-2}}{{10}^{-3}\times 0.1}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-7}\mathrm{m}$

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