Hc Verma II for Class 11 Science Physics Chapter 44 - X Rays

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.

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.

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.

Answer:

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

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.

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.

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.

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.

Answer:

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

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.

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.

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.

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.

Answer:

(c) will be halved

Cut off wavelength is given by
${\lambda }_{\min }= \frac{hc}{eV}$,
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 $\left(\lambda {\text{'}}_{\min }\right)$ is given by
$\lambda {\text{'}}_{\min }= \frac{hc}{e\left(2V\right)}$
$\Rightarrow \lambda {\text{'}}_{\min }=\frac{{\lambda }_{min}}{2}$

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.

Answer:

(c) will remain unchanged

Cut off wavelength is given by
${\lambda }_{\min }= \frac{hc}{eV}$,
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.

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.

Answer:

Using Moseley's Law,
$\sqrt{v}=a\left(Z-b\right)$,
where v = frequency of K_α X-ray
            Z = atomic number
           
$$\begin{gathered} \therefore v=a^2{\left(Z-b\right)}^2 \\ \Rightarrow {\left(Z-b\right)}^2=\frac{v}{a^2} \end{gathered}$$
This is  the equation of a parabola with some intercept on the axis, representing atomic number (Z). Hence, curve d represent this relation correctly.  

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.

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. 

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.
$\therefore V_A>V_B$

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.

            
          

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.

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.

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.

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.
            

Answer:

(b) accelerating voltage

Cutoff wavelength $\left({\lambda }_{\min }\right)$ is given by
${\lambda }_{\min } = \frac{hc}{eV}$
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.

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_L), ${\lambda }_2$ is less than ${\lambda }_1$. Hence, option (d) is not correct.

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)

Answer:

(c) the intensity remains unchanged
(d) the minimum wavelength decreases

Cutoff wavelength $\left({\lambda }_{\min }\right)$ is given by
${\lambda }_{\min } = \frac{h\mathrm{c}}{eV}$
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.

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.

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.

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
$$\begin{gathered} v=\frac{c}{{\lambda }_1}=\frac{c}{{\lambda }_2} \\ \Rightarrow \nu =\frac{c}{\lambda } \end{gathered}$$

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.

Answer:

Given:
Wavelength of the X-ray photon, λ = 0.1 nm
Speed of light, c = 3$\times$10⁸ m/s

(a) Energy of the photon $\left(E\right)$ is given by
$$\begin{gathered} E=\frac{hc}{\lambda } \\ \Rightarrow E=\frac{1242 \mathrm{eV} }{0.1 } \\ \Rightarrow E=12420 \mathrm{eV} \\ \Rightarrow E=12.42 \mathrm{keV}\approx 12.4 \mathrm{keV} \end{gathered}$$

(b) Frequency is given by
$$\begin{gathered} \nu =\frac{c}{\lambda } \\ \Rightarrow \nu =\frac{3\times {10}^8}{0.1\times {10}^{-9}} \\ \Rightarrow \nu =3\times {10}^{18} \mathrm{Hz} \end{gathered}$$

(c) Momentum of the photon $\left(P\right)$ is given by
  $$\begin{gathered} P=\frac{E}{c} \\ \Rightarrow P=\frac{12.4\times {10}^3\times 1.6\times {10}^{-19}}{3\times {10}^8} \\ \Rightarrow P=6.613\times {10}^{-24} \mathrm{kg}-\mathrm{m}/\mathrm{s}\approx 6.62\times {10}^{-24} \mathrm{g}-\mathrm{m}/\mathrm{s} \end{gathered}$$

Answer:

Given:
Energy of the X-ray, E = 6.4 keV
Distance travelled by the photon, d = 3 km = 3 $\times$ 10³ m
Time taken by the photon to cross the distance of 3 km is given by
$$\begin{gathered} t=\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{3\times {10}^3}{3\times {10}^8} \\ t={10}^{-5} \mathrm{s}=10\times {10}^{-6} \mathrm{s} \\ t=10 \mathrm{\mu s} \end{gathered}$$

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.

Answer:

Given:
Voltage of the X-ray tube, V = 30 kV
Cutoff wavelength for continuous X-rays $\left(\lambda \right)$ is given by
$$\begin{gathered} \lambda =\frac{hc}{e\mathrm{V}} \\ \mathrm{Here}, \\ h=\mathrm{Planck}\text{'}s \mathrm{constant} \\ V=\mathrm{Voltage} \mathrm{of} \mathrm{the} \mathrm{X}-\mathrm{ray} \mathrm{tube} \\ c=\mathrm{Speed} \mathrm{of} \mathrm{light} \\ \therefore \lambda =\frac{1242 \mathrm{eV}-\mathrm{nm}}{\mathrm{e}\times 30\times {10}^3} \\ \Rightarrow \lambda =414\times {10}^{-4} \mathrm{nm}=41.4 \mathrm{pm} \end{gathered}$$

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

Maximum energy of the photon $\left(E\right)$ is given by
E = $\frac{hc}{\lambda }$
$$\begin{gathered} \Rightarrow E=\frac{6.68\times {10}^{-34}\times 3\times {10}^8}{{10}^{-10}} \\ \Rightarrow E=19.89\times {10}^{-16} \\ \Rightarrow E=1.989\times {10}^{-15}\approx 2\times {10}^{-15} \mathrm{J} \end{gathered}$$

Answer:

Given:
Cutoff wavelength of the Coolidge tube, $\lambda$ = 80 pm
Energy of the electron hitting the target $\left(E\right)$ is given by
$E=\frac{hc}{\lambda }$
Here,
h = Planck's constant
c = Speed of light
$\lambda$ = Wavelength of light

$$\begin{gathered} \therefore E =\frac{1242 \mathrm{eV}-\mathrm{nm}}{80\times {10}^{-3}} \\ \Rightarrow E=\frac{1242\times {10}^{-9} \mathrm{eV}}{80\times {10}^{-12}} \\ \Rightarrow E=15.525\times {10}^3 \mathrm{eV}\approx 15.5 \mathrm{keV} \end{gathered}$$

Answer:

Let $\lambda$ be the cut off wavelength and V be the operating potential in the X-ray tube.
Then,
$\lambda =\frac{hc}{V}$
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
$$\begin{gathered} V\text{'}=\mathrm{V}+\frac{1}{100}\times \mathrm{V} \\ V\text{'}= 1.01 \mathrm{V} \end{gathered}$$

Cut off wavelength $\left(\lambda \text{'}\right)$ on increasing the operating voltage is given by
$$\begin{gathered} \lambda \text{'}=\frac{hc}{1.01V}=\frac{\lambda }{1.01} \\ \therefore \lambda \text{'}-\lambda =\frac{0.01}{1.01}\lambda \end{gathered}$$

Percentage change in the wavelength is given by
$$\begin{gathered} \frac{0.01\lambda }{1.01\times \lambda }\times 100=\frac{1}{1.01} \\ =0.9901 \end{gathered}$$
 = 1 (approx.)

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
$E=\frac{hc}{\lambda }$
Here,
h = Planck's constant
c = Speed of light
$\lambda$ = Wavelength of light

Thus, we have

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

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

$$\begin{gathered} \therefore \lambda =\frac{hc}{eV} \\ \Rightarrow \lambda V=\lambda \text{'}\mathrm{V}\text{'} \left[\because \mathrm{\lambda }\propto \frac{1}{V}\right] \\ \Rightarrow \lambda V=\left(\lambda -26\right)\times 1.5 \mathrm{V} \\ \Rightarrow 0.5\lambda =26\times 1.5 \\ \Rightarrow \lambda =26\times 3 \\ \Rightarrow \lambda =78 \mathrm{pm} \end{gathered}$$
Hence, the initial wavelength is 78$\times$10^$-$12 m.

Now, the operating voltage (V) is given by
$$\begin{gathered} V=\frac{hc}{e\lambda } \\ \Rightarrow V=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 78\times {10}^{-12}} \\ \Rightarrow V=0.15937\times {10}^5 \\ \Rightarrow V=15.9 \mathrm{kV} \end{gathered}$$

Answer:

Given:
Potential of the electron beam, V = 32 kV = 32 × 10³ V
Energy, $E=32\times {10}^3 \mathrm{eV}$
Wavelength of the X-ray photon $\left(\lambda \right)$ is given by
$\lambda =\frac{h\mathrm{c}}{E}$
Here,
h = Planck's constant
c = Speed of light         
$$\begin{gathered} \therefore \lambda =\frac{hc}{E} \\ \Rightarrow \lambda =\frac{1242 \mathrm{eVnm}}{32\times {10}^3} \\ \Rightarrow \lambda =38.8\times {10}^{-3} \mathrm{nm} \\ \Rightarrow \lambda =38.8 \mathrm{pm} \end{gathered}$$

Answer:

Given:
Potential applied to the X-ray tube, V = 40 kV
Frequency of the X-ray, v = 9.7 × 10¹⁸ Hz
Wavelength $\left(\lambda \right)$ is given by
$\lambda =\frac{hc}{eV}$
Here,
h = Planck's constant
c = Speed of light
e = 1.6 $\times$ 10^$-19$    
$$\begin{gathered} \therefore \lambda =\frac{hc}{e\mathrm{V}} \\ \Rightarrow \frac{\lambda }{c}=\frac{h}{eV} \\ \Rightarrow \frac{1}{v}=\frac{h}{eV} \left(\because v=\frac{c}{\lambda }\right) \\ \Rightarrow h=\frac{eV}{v} \\ \Rightarrow h=\frac{40\times {10}^3}{9.7\times {10}^{18}}\times e \\ \Rightarrow h=4.12\times {10}^{-15} e\mathrm{Vs} \end{gathered}$$

Answer:

Given:
Potential of the X-ray tube, V = 40 kV = 40 × 10³ V
Energy = 40 × 10³ eV
Energy utilised by the electron is given by
E =$\frac{70}{100}\times 40\times {10}^3$ = 28 $\times$ 10³ 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
$$\begin{gathered} \therefore \lambda =\frac{hc}{E} \\ \Rightarrow \lambda =\frac{1242 \mathrm{eV}-\mathrm{nm}}{28\times {10}^3 \mathrm{eV}} \\ \Rightarrow \lambda =\frac{1242\times {10}^{-9} \mathrm{eV}}{28\times {10}^3 \mathrm{eV}} \\ \Rightarrow \lambda =44.35\times {10}^{-12} \\ \Rightarrow \lambda =44.35 \mathrm{pm} \end{gathered}$$

For the second wavelength,
E = 70% (Leftover energy)
  $$\begin{gathered} =\frac{70}{100}\times (40-28){10}^3 \\ =\frac{70}{100}\times 12\times {10}^3 \\ =84\times {10}^2 \mathrm{eV} \\ \mathrm{And}, \\ \lambda =\frac{hc}{E}=\frac{1242}{8.4\times {10}^3} \\ =147.86\times {10}^{-3} \mathrm{nm} \\ =147.86 \mathrm{pm}=148 \mathrm{pm} \end{gathered}$$

For the third wavelength,
$$\begin{gathered} E=\frac{70}{100}(12-8.4)\times {10}^3 \\ =7\times 3.6\times {10}^2=25.2\times {10}^2 \mathrm{eV} \\ \mathrm{And}, \\ \lambda =\frac{hc}{E}=\frac{1242}{25.2\times {10}^2} \\ =49.2857\times {10}^{-2} \\ =493 \mathrm{pm} \end{gathered}$$

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

Answer:

Given:
Wavelength of K_α X-rays of aluminium, λ₁ = 887 pm
Frequency of X-rays of aluminium is given by
$$\begin{gathered} {\nu }_a=\frac{c}{\lambda } \\ {\nu }_a=\frac{3\times {10}^8}{887\times {10}^{-12}} \\ {\nu }_a=3.382\times {10}^{17} \\ {\nu }_a=33.82\times {10}^{16} \mathrm{Hz} \end{gathered}$$

Wavelength of K_α X-rays of zinc, ${\lambda }_2$ = 146 pm
Frequency of X-rays of zinc is given by
$$\begin{gathered} {\nu }_z=\frac{3\times {10}^8}{146\times {10}^{-12}} \\ {\nu }_z=0.02055\times {10}^{20} \\ {\nu }_z=2.055\times {10}^{18} \mathrm{Hz} \end{gathered}$$

We know
  $\sqrt{\nu }$ = a(Z − b)

For aluminium,
5.815 × 10⁸ = a(13 − b)    ...(1)

For zinc,
1.4331 × 10⁹ = a(30 − b)   ...(2)

Dividing (1) by (2)
$$\begin{gathered} \frac{13-b}{30-b}=\frac{5.815\times {10}^{-1}}{1.4331} \\ =0.4057 \\ \Rightarrow 30\times 0.4057-0.4057 b=13-b \\ \Rightarrow 12.171-0.4057 b+b=13 \\ b=\frac{0.829}{0.5943}=1.39491 \\ a=\frac{5.815\times {10}^8}{11.33} \\ =0.51323\times {10}^8=5\times {10}^7 \end{gathered}$$

For Fe,
Frequency $\left(\nu \text{'}\right)$ is given by
$\nu$' = 5× 10⁷ (26 − 1.39)
  = 5 × 24.61 × 10⁷
  = 123.05 × 10⁷

$\nu$' = $\frac{c}{\lambda }$
Here, c =  speed of light
          $\lambda$ = Wavelength of light
$$\begin{gathered} \therefore \frac{c}{\lambda }=5141.3\times {10}^{14} \\ \Rightarrow \lambda =\frac{3\times {10}^8}{5141.3\times {10}^{14}} \\ =0.000198\times {10}^{-5} \mathrm{m} \\ =198\times {10}^{-12}=198 \mathrm{pm} \end{gathered}$$

Answer:

Given:
Energy of K_α X-ray, E = 3.69 keV = 3690 eV

Wavelength $\left(\lambda \right)$ is given by
$$\begin{gathered} \lambda =\frac{hc}{E} \\ \lambda =\frac{1242}{3690} \\ \lambda =0.33658 \mathrm{nm} \\ \Rightarrow \lambda =0.34\times {10}^{-9} \mathrm{m} \end{gathered}$$   

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,
$$\begin{gathered} \sqrt{\frac{3\times {10}^8}{0.34\times {10}^{-9}}}=5\times {10}^7(Z-1.39) \\ \Rightarrow \sqrt{8.82\times {10}^{17}}=5\times {10}^7(Z-1.39) \\ \Rightarrow 9.39\times {10}^8=5\times {10}^7(Z-1.39) \\ \Rightarrow \frac{93.9}{5}=Z-1.39 \\ \Rightarrow Z=\frac{93.9}{5}+1.39 \\ =20.17=20 \end{gathered}$$
So, the element is calcium.

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:

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,
$\sqrt{v}=a(\mathrm{Z}-b)$
Here, v = frequency of X-ray
         Z = atomic number of the element
$$\begin{gathered} \sqrt{v}=a(\mathrm{Z}-b) \\ \Rightarrow \sqrt{v}=a(57-1) \\ =a\times 56 ...\left(1\right) \\ \mathrm{For} \mathrm{C}u \left(29\right), \\ \sqrt{1.88\times {10}^{18}}=a(29-1)=28 a ...\left(2\right) \end{gathered}$$
Dividing (1) and (2)
$$\begin{gathered} \sqrt{\frac{v}{1.88\times {10}^{18}}}=\frac{a\times 56}{a\times 28}=2 \\ \Rightarrow v=1.88\times {10}^{18}\times (2{)}^2 \\ =4\times 1.88\times {10}^{18} \\ =7.52\times {10}^{18} \mathrm{Hz} \end{gathered}$$

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_αX-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),

$$\begin{gathered} K_{\mathrm{\alpha }}-K_{\mathrm{\beta }}=E_{\mathrm{M}}-E_{\mathrm{L}}=-K_L \\ \mathrm{or} K_L=K_{\mathrm{\beta }}-K_{\mathrm{\alpha }} \\ K_L=\frac{3\times {10}^8}{0.63\times {10}^{-10}}-\frac{3\times {10}^8}{0.71\times {10}^{-10}} \\ K_L=4.761\times {10}^{-18}-4.225\times {10}^{18} \\ K_L=0.536\times {10}^{18} \mathrm{Hz} \\ \mathrm{Again}, \mathrm{\lambda }=\frac{3\times {10}^8}{0.536\times {10}^{-18}} \\ \Rightarrow \mathrm{\lambda }=5.6\times {10}^{-10}=5.6 \mathrm{\AA } \end{gathered}$$

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
$$\begin{gathered} E_1=\frac{1242}{21.3\times {10}^{-3}} \\ =58.309\times {10}^{3 }\mathrm{eV} \end{gathered}$$

Energy of L_α X-ray $\left(E_2\right)$ is given by
$$\begin{gathered} E_2=\frac{1242}{141\times {10}^{-5}} \\ =8.8085\times {10}^3 \mathrm{eV} \end{gathered}$$

Energy of K_β X-ray $\left(E_3\right)$ will be
$$\begin{gathered} E_3=E_1+E_2 \\ {E}_3=(58.309+8.809)\times {10}^3 \mathrm{eV} \\ E_3=67.118\times {10}^3 \mathrm{eV} \end{gathered}$$
Wavelength of K_β X-ray $\left(\lambda \right)$ is given by
$$\begin{gathered} \mathrm{\lambda }=\frac{hc}{E_3}=\frac{1242}{67.118\times {10}^3} \\ =18.5\times {10}^{-3 }\mathrm{nm}=18.5 \mathrm{pm} \end{gathered}$$

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 f₀ be the frequency of K_β X-ray.
Let f₁ 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

 $$\begin{gathered} f=\frac{E_{\mathrm{K}}-E_{\mathrm{L}}}{h} \\ f=\frac{(25.31-3.56)}{6.63\times {10}^{-34}}\times 1.6\times {10}^{-19}\times {10}^3 \\ f=\frac{21.75\times {10}^3\times {10}^{15}}{6.67} \\ f=5.25\times {10}^{18 }\mathrm{Hz} \\ {\mathrm{K}}_{\mathrm{\beta }}=E_{\mathrm{K}}-E_{\mathrm{M}}=h{f}_0 \\ \Rightarrow f_0=\frac{E_{\mathrm{K}}-E_{\mathrm{M}}}{h} \\ \Rightarrow f_0=\frac{(25.31-0.53)}{6.67\times {10}^{-34}}\times {10}^3\times 1.6\times {10}^{-19} \\ \Rightarrow f_0=5.985\times {10}^{18} \mathrm{Hz} \end{gathered}$$

$$\begin{gathered} K_L=E_L-E_M=h{f}_1 \\ {f}_1=\frac{E_L-E_M}{h} \\ {f}_1= \frac{3.56-0.530}{6.63\times {10}^{-34}}\times {10}^3\times 1.6\times {10}^{-19} \\ {f}_1= 7.32\times {10}^{17} \mathrm{Hz} \end{gathered}$$

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.

Answer:

Given:
V = 30 KV
i = 10mA

1% of T_KE (total kinetic energy) = X-ray
i = ne
$$\begin{gathered} \mathrm{or} n=\frac{{10}^{-2}}{1.6\times {10}^{-19}} \\ =0.625\times {10}^{17} (\mathrm{number} \mathrm{of} \mathrm{electrons}) \end{gathered}$$


KE of one electron = eV
$$\begin{gathered} =1.6\times {10}^{-19}\times 40\times {10}^3 \\ =6.4\times {10}^{-15}\mathrm{J} \\ {\mathrm{T}}_{\mathrm{KE}}=0.625\times 6.4\times {10}^{17}\times {10}^{-15} \\ =4\times {10}^2\mathrm{J} \end{gathered}$$
(b) Heat produced in target per second = 400 − 4 = 396 J

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
$$\begin{gathered} \therefore P=\frac{ne\mathrm{V}}{t} \\ \Rightarrow 200 =\frac{neV}{t} \\ \Rightarrow (ne/t)\mathrm{V}=200 \\ \Rightarrow I=\frac{200}{V} \\ \Rightarrow I=\frac{200}{20\times {10}^3}=10 \mathrm{mA} \end{gathered}$$

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

Answer:

Given:
f = (25 × 10¹⁴ Hz) (Z − 1)^{2
 $$\begin{gathered} \mathrm{or} \frac{c}{\lambda }=25\times {10}^{14}(\mathrm{Z}-1{)}^2 \\ (\mathrm{a}) \frac{3\times {10}^8}{78.9\times {10}^{-12}\times 25\times {10}^{14}}=(\mathrm{Z}-1{)}^2 \\ \mathrm{or} (\mathrm{Z}-1{)}^2=.0.001520\times {10}^6=1520 \\ \Rightarrow \mathrm{Z}-1=38.98 \\ \mathrm{or} \mathrm{Z}=39.98=40 \\ \mathrm{Thus}, \mathrm{it} \mathrm{is} (\mathrm{Zr}). \\ (\mathrm{b}) \frac{3\times {10}^8}{146\times {10}^{-12}\times 25\times {10}^{14}}=(\mathrm{Z}-1{)}^2 \\ \mathrm{or} (\mathrm{Z}-1{)}^2=0.00082219\times {10}^6 \\ \mathrm{or} \mathrm{Z}-1=28.669 \\ \mathrm{or} \mathrm{Z}=29.669=30 \\ \mathrm{Thus}, \mathrm{it} \mathrm{is} (\mathrm{Zn}). \\ (\mathrm{c}) \frac{3\times {10}^8}{158\times {10}^{-12}\times 25\times {10}^4}=(Z-1{)}^2 \\ \mathrm{or} (\mathrm{Z}-1{)}^2=0.0007594\times {10}^4 \\ \mathrm{or} \mathrm{Z}-1=27.5559 \\ \mathrm{or} \mathrm{Z}=28.5589=29 \\ \mathrm{Thus}, \mathrm{it} \mathrm{is} (\mathrm{Cu}). \\ (\mathrm{d}) \frac{3\times {10}^8}{198\times {10}^{-12}\times 25\times {10}^4}=(\mathrm{Z}-1{)}^2 \\ \mathrm{or} (\mathrm{Z}-1{)}^2=0.000606\times {10}^4 \\ \mathrm{or} \mathrm{Z}-1=24.6162 \\ \mathrm{or} \mathrm{Z}=25.6162=26 \\ \mathrm{Thus}, \mathrm{it} \mathrm{is} (\mathrm{Fe}). \end{gathered}$$}

Answer:

Here, energy of photon = E
E = 6.4 KeV = 6.4 × 10³ eV
Momentum of photon
$$\begin{gathered} =\frac{\mathrm{E}}{c}=\frac{6.4\times {10}^3}{3\times {10}^8} \\ =3.41\times {10}^{-24}\mathrm{kgm}/\sec \end{gathered}$$
According to collision theory,
Momentum of photon = Momentum of atom
∴ Momentum of atom, P = 3.41 × 10⁻²⁴ kgm/sec

 Recoil KE of atom = P²/2M
$=\frac{(3.41\times {10}^{-24}{)}^{2}e\mathrm{V}}{\left(20\right)9.3\times {10}^{-26}\times 1.6\times {10}^{-19}}=3.9e\mathrm{V}$
[1 Joule = 1.6 × 10⁻¹⁹ eV]

Answer:

V₀ - Stopping Potential
K - Potential difference across X-ray tube
λ - Wavelength
λ - Cut difference Wavelength
$$\begin{gathered} e{\mathrm{V}}_0=hf-h{f}_0 \\ \lambda =\frac{hc}{e\mathrm{V}} \\ e{\mathrm{V}}_0=\frac{hc}{\lambda } \\ \mathrm{or} \mathrm{V\lambda }=\frac{hc}{e} \\ \mathrm{or} {\mathrm{V}}_0\mathrm{\lambda }=\frac{hc}{e} \end{gathered}$$
Here, the slopes are same.
i.e. V₀λ = Vλ
$$\begin{gathered} \frac{hc}{e}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}} \\ =1.242\times {10}^{-6} \mathrm{Vm} \end{gathered}$$

Answer:

Given:
$$\begin{gathered} \lambda =10 \mathrm{pm}=100\times {10}^{-12}\mathrm{m} \\ \mathrm{D}=40 \mathrm{cm}=40\times {10}^{-2}\mathrm{m} \\ \mathrm{\beta }=0.1 \mathrm{mm}=0.1\times {10}^{-3}\mathrm{m} \\ \mathrm{\beta }=\frac{\mathrm{\lambda D}}{d} \\ d=\frac{\mathrm{\lambda D}}{\mathrm{\beta }} \\ =\frac{100\times {10}^{-12}\times 40\times {10}^{-2}}{{10}^{-3}\times 0.1} \\ =4\times {10}^{-7}\mathrm{m} \end{gathered}$$