5 Marks Questions

Question 15 Marks

The _{$\text{K}_\beta$} X-rays from certain elements are given below. Draw a Moseley-type plot of $\sqrt{\text{v}}$ versus Z for $\text{K}_\beta$ radiation.

Element	Ne	P	Ca	Mn	Zn	Br
Energy (keV)	0.858	2.14	4.02	6.51	9.57	13.3

Answer

$\text{K}_\text{B}$ radiation is when the e jumps from

n = 3 to n = 1 (here n is principal quantum no)

$\Delta\text{E}=\text{hv}=\text{Rhc(z-h)}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$

$\sqrt{\text{v}}=\sqrt{\frac{9\text{RC}}{8}}(\text{z}-\text{h})$

$\therefore\sqrt{\text{v}}\propto\text{z}$

Second method :

We can directly get value of v b

hv = Energy

$\Rightarrow\text{v}=\frac{\text{Energy(in Kev)}}{\text{h}}$

This we have to find out $\sqrt{\text{v}}$ and draw the same graph as above.

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Question 25 Marks

Use Moseley's law with b = 1 to find the frequency of the $\text{K}_\alpha$ X-ray of La(Z = 57) if the frequency of the $\text{K}_\alpha$ X-ray of Cu(Z = 29) is known to be 1.88 × 10¹⁸Hz.

Answer

b = 1

For $\propto\text{a}(57)$

$\sqrt{\text{v}}=\text{a}(\text{Z-b})$

$\Rightarrow\sqrt{\text{v}}=\text{a}(57-1)=\text{a}\times56\ ...(1)$

For Cu(29)

$\sqrt{1.88\times10^{78}}=\text{a}(29-1)=28\text{a}\ ...(2)$

dividing (1) and (2)

$\sqrt{\frac{\text{V}}{1.88\times10^{18}}}=\frac{\text{a}\times56}{\text{a}\times28}=2$

$\Rightarrow\text{v}=1.88\times10^{18}(2)^2$

$=4\times1.88\times10^{18}=7.52\times10^8\text{Hz}$

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Question 35 Marks

The short-wavelength limit shifts by 26pm 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

Give

$\lambda'=\lambda-26\text{pm},$

$\text{V}'=1.5\text{V}$

Now $\lambda=\frac{\text{hc}}{\text{ev}},$

$\lambda'=\frac{\text{hc}}{\text{ev}'}$

Or $\lambda\text{V}=\lambda'\text{V}'$

$\Rightarrow\lambda\text{V}=\Big(\lambda-26\times10^{-12}\Big)\times1.5\text{V}$

$\Rightarrow\lambda=1.5\lambda-1.5\times26\times10^{-12}$

$\Rightarrow\lambda=\frac{39\times10^{-12}}{0.5}$

$=78\times10^{-12}\text{m}$

$\text{V}=\frac{\text{hc}}{\text{e}\lambda}$

$=\frac{6.63\times3\times10^{-34}\times10^8}{1.6\times10^{-19}\times78\times10^{-12}}$

$=0.15937\times10^5=15.93\times10^3\text{V}$

$15.93\text{KV}.$

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Question 45 Marks

An X-ray tube operates at 40kV. 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

V = 40KV = 40 × 10³V

Energy = 40 × 10³eV

Energy utilized $=\frac{70}{100}\times40\times10^3$

$=28\times10^3\text{eV}$

$\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242-\text{ev-nm}}{28\times10^3\text{ev}}$

$\Rightarrow44.35\times10^{-3}\text{nm}=44.35\text{pm}$

For other wavelengths,

E = 70% (left over energy)

$=\frac{70}{100}\times(40-28)10^3=84\times10^2$

$\lambda'=\frac{\text{hc}}{\text{E}}=\frac{1242}{8.4\times10^3}$

$=147.86\times10^{-3}\text{nm}$

$=147.86\text{pm}=148\text{pm}$

For third wavelengths,

$\text{E}=\frac{70}{100}=(12-8.4)\times10^3$

$=7\times3.6\times10^2=25.2\times10^2$

$\lambda'=\frac{\text{hc}}{\text{E}}=\frac{1242}{25.2\times10^2}$

$=49.2857\times10^{-2}\text{nm}=493\text{pm}$

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Question 55 Marks

The $\text{K}_\alpha$ and $\text{K}_\beta$ X-rays of molybdenum have wavelengths 0.71A and 0.63A respectively. Find the wavelength of $\text{L}_\alpha$ X-ray of molybdenum.

Answer

$\text{K}_\alpha=\text{E}_\text{K}-\text{E}_\text{L}\ ...(1)\ \lambda\text{K}_\beta=0.71\mathring{\text{A}}$

$\text{K}_\beta=\text{E}_\text{K}-\text{E}_\text{M}\ ...(2)\ \lambda\text{K}_\beta=63\mathring{\text{A}}$

$\text{L}_\alpha=\text{E}_\text{L}-\text{E}_\text{L}-\text{E}_\text{M}\ ...(3)$

Subtracting (2) from (1)

$\text{K}_\alpha-\text{K}_\beta=\text{E}_\text{M}-\text{E}_\text{L}=-\text{L}_\alpha$

Or, $\text{L}_\alpha=\text{K}_\beta-\text{K}_\alpha$

$=\frac{3\times10^8}{0.63\times10^{-10}}-\frac{3\times10^8}{0.71\times10^{-10}}$

$=4.761\times10^{18}-4.225\times10^{18}$

$=0.536\times10^{18}\text{Hz}$

Again $\lambda=\frac{3\times10^8}{0.536\times10^{18}}$

$=5.6\times10^{-10}=5.6\mathring{\text{A}}$

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Question 65 Marks

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 $\text{K}_\alpha$ intensities are detected, list the elements present in the sample from the plot. Use Moseley's equation v - (25 × 10¹⁴Hz)(Z - 1)².

Answer

Given

v = (25 × 10¹⁴Hz) (Z -1)²

$\frac{\text{C}}{\lambda}=25\times10^{14}(\text{Z}-1)^2$

$\frac{\text{3}\times10^8}{78.9\times10^{-12}\times25\times10^{14}}=(\text{Z}-1)^2$

$(\text{Z}-1)^2=38.98$

$\text{Z}=39.98=40$ It is (Zr)

$\frac{3\times10^8}{146\times10^{-12}\times25\times10^{14}}=(\text{Z}-1)^2$

$(\text{Z}-1)^2=0.0008219\times10^6$

$\Rightarrow\text{Z}-1=28.669 $

$\text{Z}=29.669=30$ It is (Zn).

$\frac{3\times10^8}{158\times10^{-12}\times25\times10^{14}}=(\text{Z}-1)^2$

$(\text{Z}-1)^2=0.0007594\times10^6$

$\Rightarrow\text{Z}-1=27.5589$

$\text{Z}=28.5589=29$ It is (Cu).

$\frac{3\times10^8}{19\times10^{-12}\times25\times10^{14}}=(\text{Z}-1)^2$

$(\text{Z}-1)^2=0.000606\times10^6$

$\Rightarrow\text{Z}-1=24.6182 $

$\text{Z}=25.6182=26$ It is (Fe).

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Question 75 Marks

The wavelengths of $\text{K}_\alpha$ and $\text{L}_\alpha$ X-rays of a material are 21.3pm and 141pm respectively. Find the wavelength of $\text{K}_\beta$ X-ray of the material.

Answer

$\text{E}_1=\frac{1242}{21.3\times10^{-3}}=58.309\times10^3\text{eV}$

$\text{E}_2=\frac{1242}{141\times10^{-3}}=8.8085\times10^3\text{eV}$

$\text{E}_3=\text{E}_1+\text{E}_2$

$\Rightarrow(58.309+8.809)\text{ev}=67.118\times10^3\text{ev}$

$\lambda=\frac{\text{hc}}{\text{E}_3}=\frac{1242}{67.118\times10^3}$

$=18.5\times10^{-3}\text{nm}=18.5\text{pm}$

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Question 85 Marks

Is it possible that in a Coolidge tube characteristic $\text{L}_\alpha$ X-rays are emitted but not $\text{K}_\alpha$ X-rays?

Answer

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

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Question 95 Marks

The energy of a silver atom with a vacancy in K shell is 25.31keV, in L shell is 3.56keV and in M shell is 0.530keV higher than the energy of the atom with no vacancy. Find the frequency of $\text{K}_\alpha\text{K}_\beta$ and $\text{L}_\alpha$ X-rays of silver.

Answer

Given:

Energy of electron in the K shell, E_k = 25.31keV

Energy of electron in the L shell, E_L = 3.56keV

Energy of electron in the M shell, E_M = 0.530keV

Let f be the frequency of $\text{K}_\alpha$ X-ray and f₀ be the frequency of $\text{K}_\beta$ X-ray.

Let f₁ be the frequency of $\text{L}_\alpha$ X-rays of silver.

$\therefore\text{K}_\alpha=\text{E}_\text{K}-\text{E}_\text{L}=\text{hf}$

Here,

h = Planck constant

f = frequency of $\text{K}_\alpha$ X-ray

$\text{f}=\frac{\text{E}_\text{K}-\text{E}_\text{L}}{\text{h}}$

$\text{f}=\frac{(25.31-3.56)}{6.63\times10^{-34}}\times1.6\times10^{-19}\times10^3$

$\text{f}=\frac{21.75\times10^3\times10^{15}}{6.67}$

$\text{f}=5.25\times10^{18}\text{Hz}$

$\text{K}_\beta=\text{E}_\text{K}-\text{E}_\text{M}=\text{hf}_0$

$\Rightarrow\text{f}_0=\frac{\text{E}_\text{K}-\text{E}_\text{M}}{\text{h}}$

$\Rightarrow\text{f}_0=\frac{(25.31-0.53)}{6.67\times10^{-34}}\times10^3\times1.6\times10^{-19}$

$\Rightarrow\text{f}_0=5.985\times10^{18}\text{Hz}$

$\text{K}_\text{L}=\text{E}_\text{L}-\text{E}_\text{M}=\text{hf}_1$

$\text{f}_1=\frac{\text{E}_\text{L}-\text{E}_\text{M}}{\text{h}}$

$\text{f}_1=\frac{3.56-0.530}{6.63\times10^{-34}}\times10^3\times1.6\times10^{-19}$

$\text{f}_1=7.32\times10^{17}\text{Hz}$

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Question 105 Marks

Can $\text{L}_\alpha$ X-ray of one material have shorter wavelength than $\text{K}_\alpha$ X-ray of another?

Answer

An $\text{L}\alpha$ X-ray is emitted when an electron jumps from the M to the L shell, and a $\text{K}_\alpha$ 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 $\text{L}\alpha$ X-ray has higher wavelength than a $\text{K}_\alpha$ X-ray for the same material.

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Question 115 Marks

The $\text{K}_\alpha$ X-rays of aluminium (Z = 13) and zinc (Z = 30) have wavelengths 887pm and 146pm respectively. Use Moseley's law $\sqrt{\text{v}}=\text{a}(\text{z-b})$ to find the wavelengths of the $\text{K}_\alpha$ X-ray of iron (Z = 26).

Answer

$\lambda_1=887\text{pm}$

$\text{v}=\frac{\text{C}}{\lambda}=\frac{3\times10^8}{887\times10^{-12}}$

$=3.383\times10^7=33.82\times10^{16}$

$=5.815\times10^8$

$\lambda_2=146\text{pm}$

$\text{v}=\frac{3\times10^8}{146\times10^{-12}}$

$=0.02054\times10^{20}=2.054\times10^{18}$

$=1.4331\times10^9$

We know, $\sqrt{\text{v}}=\text{a}(\text{z-b})$

$\Rightarrow\frac{\sqrt{5.815\times10^8}\ =\text{ a}(13-\text{b})}{\sqrt{1.4331\times10^9}\ =\text{ a}(30-\text{b})}$

$\Rightarrow\frac{13-\text{b}}{30-\text{b}}=\frac{5.815\times10^{-1}}{1.4331}=0.4057$

$\Rightarrow30\times0.4057-0.4057\text{b}$

$=13-\text{b}$

$\Rightarrow12.171-0.4.57\text{b}+\text{b}=13$

$\Rightarrow\text{b}=\frac{0.829}{0.5943}=1.39491$

$\Rightarrow\text{a}=\frac{5.815\times10^8}{11.33}=0.51323\times10^8$

$=5\times10^7$

For ‘Fe’,

$\sqrt{\text{v}}=5\times10^7(26-1.39)$

$=5\times24.61\times10^7=123.05\times10^7$

$\frac{\text{C}}{\lambda}=15141.3\times10^{14}$

$=\lambda=\frac{3\times10^8}{15141.3\times10^{14}}$

$=0.000198\times10^{-6}\text{m}=198\times10^{-12}$

$=198\text{pm}$

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Question 125 Marks

A certain element emits $\text{K}_\alpha$ X-ray of energy 3.69keV. Use the data from the previous problem to identify the element.

Answer

$\text{E}=3.69\text{Kev}=3690\text{eV}$

$\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242}{3690}=0.33658\text{nm}$

$\sqrt{\frac{\text{C}}{\lambda}}\text{a}(\text{z}-\text{b})$

$\text{a}=5\times10^7\sqrt{\text{Hz}}$

$\text{b}=1.37$ (from previous problem)

$\sqrt{\frac{3\times10^8}{0.34\times10^{-9}}}$

$=5\times10^7(\text{Z}-1.37)$

$\Rightarrow\sqrt8.82\times10^{-17}=5\times10^7({\text{Z}-1.37)}$

$\Rightarrow9.39\times10^8=5\times10^7(\text{Z}-1.37)$

$\Rightarrow\frac{93.9}{5}=\text{Z}-1.37$

$\Rightarrow\text{Z}=20.15=20$

$\therefore$ The element is calcium.

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Question 135 Marks

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.

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