# X-rays

## Concept Of Physics

### H C Verma

1   Find the energy, the frequency and the momentum of an X-ray photon of wavelength $0\cdot10\ nm$.

##### Solution :

$\lambda = 0.1\ nm$ $\\$ a) Energy $= \dfrac{hc}{\lambda} = \dfrac{1242\ ev.nm}{0.1\ nm}$ $\\$ $= 12420\ ev = 12.42\ Kev = 12.4\ kev$.

b) Frequency $= \dfrac{C}{\lambda} = \dfrac{3 \times 10^{8}}{0.1 \times 10^{-9}} = \dfrac{3 \times 10^8}{10^{-10}} = 3 \times 10^{18} Hz$ $\\$ c) Momentum $= \dfrac{E}{C} = \dfrac{12.4 \times 10^3 \times 1.6 \times 10^{-19}}{3 \times 10^8} = 6.613 \times10^{–24}\ kg-m/s = 6.62 \times 10^{–24}\ kg-m/s.$

2   Iron emits $K_a$ $X$-ray of energy $6\cdot4\ keV$ and calcium emits $K_a$ $X$-ray of energy $3\cdot69\ keV$. Calculate the times taken by an iron $K_a$ photon and a calcium Ka photon to cross through a distance of $3\ km.$

##### Solution :

Distance $= 3\ km = 3 \times 103\ m$ $\\$ $C = 3 \times 108\ m/s$ $\\$ $t =\dfrac{Dist}{Speed} = \dfrac{3 \times 10^3}{3 \times 10^8} =10^{-5}\ sec.$ $\\$ $\Rightarrow 10 \times 10^{–8}\ sec = 10\ \mu{s}$ in both case.

3   Find the cut off wavelength for the continuous $X$-rays coming from an $X$-ray tube operating at $30\ kV.$

##### Solution :

$V = 30\ KV$ $\lambda = \dfrac{hc}{E} = \dfrac{hc}{x eV} = \dfrac{1242\ ev\ -\ nm}{e \times 30 \times 10^3} = 414 \times 10^{–4}\ nm = 41.4\ Pm.$

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 ?

##### Solution :

$\lambda = 0.10\ nm = 10^{–10}\ m ; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = 6.63 \times 10^{–34}\ J-s$ $C = 3 \times 10^8\ m/s ; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = 1.6 \times 10^{–19}\ C$

$\lambda_{min} = \dfrac{hc}{eV} \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \dfrac{hc}{e\lambda}$ $\\$ $= \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 10^{-10}} = 12.43 \times 10^3\ V = 12.4\ KV.$ $\\$ Max. Energy $= \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{10^{-10}} = 19.89 \times 10^{–18} = 1.989 \times 106{–15} = 2 \times 10^{–15}\ J$

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.

##### Solution :

$\lambda = 80\ pm,\ E = \dfrac{hc}{\lambda} = {1242}{80 \times 10^{-3}} = 15.525 \times 10^3\ eV = 15.5\ KeV$

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

##### Solution :

We know $\lambda = \dfrac{hc}{V}$ $\\$ Now $\lambda = \dfrac{hc}{1.01V} = \dfrac{\lambda}{1.01}$ $\\$ $\lambda - \lambda' = \dfrac{0.01}{1.01} \lambda$ $\\$ % change of wave length $= \dfrac{0.01 \times \lambda}{1.01 \times \lambda} \times 100 = \dfrac{1}{1.01} = 0.9900 = 1$%

7   The distance between the cathode (filament) and the target in an $X$-ray tube is $1\cdot5\ m$. If the cut off wavelength is $30\ pm$, find the electric field between the cathode and the target.

##### Solution :

$d = 1.5\ m,\ \lambda = 30\ pm = 30 \times 10^{–3}\ nm$ $\\$ $E = {hc}{\lambda} = \dfrac{1242}{30 \times 10^{-3}} = 41.4 \times 10^3\ eV$ $\\$ Electric field $= \dfrac{V}{d} = \dfrac{41.4 \times 10^3}{1.5} = 27.6 \times 10^3\ V/m = 27.6\ KV/m.$

8   The short-wavelength limit shifts by $26\ pm$ when the operating voltage in an $X$-ray tube is increased to $1\cdot5$ times the original value. What was the original value of the operating voltage ?

##### Solution :

Given $\lambda{'} = \lambda – 26\ pm, V' = 1.5 V$ $\\$ Now, $\lambda = \dfrac{hc}{ev}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda{'} = \dfrac{hc}{ev'}$ $\\$ or $\lambda{V} = \lambda'V'$ $\\$ $\Rightarrow \lambda{V} = (\lambda – 26 \times 10^{–12}) \times 1.5\ V$ $\\$ $\Rightarrow \lambda = 1.5 \lambda – 1.5 \times 26 \times 10^{–12}$ $\\$ $\Rightarrow \lambda = \dfrac{39 \times 10^{-12}}{0.5} = 78 \times 10^{–12}\ m$ $\\$ $V = \dfrac{hc}{e\lambda} = {6.63 \times 3 \times 10^{-34} \times 10^8}{1.6 \times 10^{-19} \times 78 \times 10^{-12}} = 0.15937 \times 10^5 = 15.93 \times 10^3\ V = 15.93\ KV.$

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 ?

##### Solution :

$V = 32\ KV = 32 \times 10^3\ V$ $\\$ When accelerated through $32\ KV$ $\\$ $E = 32 \times 10^3\ eV$ $\\$ $\lambda = \dfrac{hc}{E} \dfrac{1242}{32 \times 10^3} = 38.8 \times 10^{–3}\ nm = 38.8\ pm.$

10   When $40\ kV$ is applied across an $X$-ray tube, $X$-ray is obtained with a maximum frequency of $9\cdot7 \times 10^{16}\ Hz$. Calculate the value of Planck constant from these data.

##### Solution :

$\lambda = \dfrac{hc}{eV} ; V = 40\ kV,\ f = 9.7 \times 10^{18}\ Hz$ $\\$ or, $\dfrac{h}{c} = \dfrac{h}{eV} ;\ or,\ \dfrac{i}{f} = \dfrac{h}{eV} ;\ or\ h = \dfrac{eV}{f} V - s$ $\\$ $= \dfrac{eV}{f} V - s = \dfrac{40 \times 10^3}{9.7 \times 10^{18}} = 4.12 \times 10^{–15}\ eV-s.$

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 three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

##### Solution :

$V = 40\ KV = 40 \times 10^3\ V$ $\\$ Energy $= 40 \times 10^3\ eV$ $\\$ Energy utilized $= \dfrac{70}{100} \times 40 \times 10^3 = 28 \times 10^3\ eV$ $\\$ $\lambda = \dfrac{hc}{E} = \dfrac{1242\ ev\ nm}{28 \times 10^3\ ev} \Rightarrow 44.35 \times 10^{–3}\ nm = 44.35\ pm.$ $\\$ For other wavelengths, $\\$ $E = 70$% (left over energy) $= \dfrac{70}{100} \times (40 - 28)10^3 = 84 \times 10^2.$ $\\$ $\lambda{'} = \dfrac{hc}{E} = \dfrac{1242}{8.4 \times 10^3} = 147.86 \times 10^{–3}\ nm = 147.86\ pm = 148\ pm.$ $\\$ For third wavelength, $\\$ $E = \dfrac{70}{100} = (12 – 8.4) \times 10^3 = 7 \times 3.6 \times 10^2 = 25.2 \times 10^2$ $\\$ $\lambda{'} = \dfrac{hc}{E} \dfrac{1242}{25.2 \times 10^2} = 49.2857 \times 10^{–2}\ nm = 493\ pm.$

12   The wavelength of $K_a$ $X$-ray of tungsten is $21\cdot3$ pm. It takes $11\cdot3\ keV$ 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 $K_0$ $X$-ray ?

##### Solution :

$K_{\\ambda} = 21.3 \times 10{–12}\ pm, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ Now, $E_K – E_L = \dfrac{1242}{21.3 \times 10^{-3}} = 58.309 kev$ $\\$ $E_L = 11.3\ kev, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_K = 58.309 + 11.3 = 69.609\ kev$ $\\$ Now, $Ve = 69.609\ KeV,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or\ V = 69.609\ KV$.

13   The $K_p$ $X$-ray of argon has a wavelength of $0\cdot36\ nm$. 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.

##### Solution :

$\lambda = 0.36\ nm$ $\\$ $E = \dfrac{1242}{0.36} = 3450\ eV\ (E_M – E_K)$ $\\$ Energy needed to ionize an organ atom $= 16\ eV$ $\\$ Energy needed to knock out an electron from $K$-shell $\\$ $= (3450 + 16)\ eV = 3466\ eV = 3.466\ KeV.$

14   The $K_0\ X$-rays of aluminium $(Z = 13)$ and zinc $(Z = 30)$ have wavelengths $887$ pm and $146$ pm respectively. Use Moseley's law $\sqrt{v} = a(Z - b)$ to find the wavelength of the $K_a$ $X$-ray of iron $(Z\ -\ 26)$.

##### Solution :

$\lambda_1 = 887\ pm$ $\\$ $v = {C}{\lambda} = \dfrac{3 \times 10^8}{887 \times 10^{-12}} = 3.382 \times 10^7 = 33.82 \times 10^{16} = 5.815 \times 10^8$ $\\$ $\lambda_{2} = 146\ pm$ $\\$ $v = \dfrac{3 \times 10^8}{146 \times 10^{-12}} = 0.02054 \times 10^{20} = 2.054 \times 10^{18} = 1.4331 \times 10^9.$ $\\$

We know, $\sqrt{v} = a(z \times b)$ $\\$ $\Rightarrow \dfrac{\sqrt{5.815 \times 10^8} = a(13 - b)}{\sqrt{1.4331 \times 10^9} = a(30 - b)}$ $\\$ $\Rightarrow \dfrac{13 - b }{30 - b} = \dfrac{5.815 \times 10^{-1}}{1.4331} = 0.4057$ $\\$ $\Rightarrow 30 \times 0.4057 – 0.4057\ b = 13 – b$ $\\$ $\Rightarrow 12.171 – 0.4.57 b + b = 13$ $\\$ $\Rightarrow b = \dfrac{0.829}{0.5943} = 1.39491$ $\\$ $\Rightarrow a = \dfrac{5.815 \times 10^8}{11.33}= 0.51323 \times 10^8 = 5 \times 10^7$. $\\$ For ‘Fe’, $\\$ $\sqrt{v} = 5 \times 10^7 (26 – 1.39) = 5 \times 24.61 \times 10^7 = 123.05 \times 10^7$ $\\$ $\dfrac{c}{\lambda} = 15141.3 \times 10^{14}$ $\\$ $= \lambda = \dfrac{3 \times 10^8}{15141.3 \times 10^{14}} = 0.000198 \times 10{–6}\ m = 198 \times 10^{–12} = 198\ pm.$

15   A certain element emits $K_a\ X$-ray of energy $3\cdot69\ keV$. Use the data from the previous problem to identify the element.

##### Solution :

$E = 3.69\ kev = 3690\ eV$ $\\$ $\lambda = \dfrac{hc}{E} = \dfrac{1242}{3690} = 0.33658\ nm$ $\\$ $\sqrt{\dfrac{c}{\lambda}} = a(z – b);\ a = 5 \times 10^7\ Hz ,\ b = 1.37$ (from previous problem) $\\$ $\sqrt{\dfrac{3 \times 10^8}{0.34 \times 10^{-9}}} = 5 \times 10^7 (Z - 1.37) \Rightarrow \sqrt{8.82 \times 10^{17}} = 5 \times 10^7 (Z - 1.37)$ $\\$ $\Rightarrow 9.39 \times 10^8 = 5 \times 10^7 (Z – 1.37) \Rightarrow \dfrac{93.9}{5} = Z – 1.37$ $\\$ $\Rightarrow Z = 20.15 = 20$ $\\$ $\therefore$ The element is calcium.

16   The Kp X-rays from certain elements are given below. Draw a Moseley-type plot of /v versus Z for Kp radiation.$\\$ Element $: \ \ \ \ \ \ \ \ Ne\ \ \ \ P\ \ \ \ Ca\ \ \ \ Mn\ \ \ \ Zn\ \ \ \ Br\ \ \ \$ $\\$ Energy $(keV) : \ \ 0\cdot858 \ \ \ \ 2\cdot14\ \ \ \ 4\cdot02\ \ \ \ 6\cdot51\ \ \ \ 9\cdot57\ \ \ \ 13\cdot3.$

##### Solution :

$K_B$ radiation is when the e jumps from $n = 3$ to $n = 1$ (here $n$ is principal quantum no) $\\$ $\Delta{E} = h\nu = Rhc (z - h)^2 \Big(\dfrac{1}{2^2} - \dfrac{1}{3^2}\Big)$ $\\$ $\Rightarrow \sqrt{v} = \sqrt{\dfrac{9RC}{8}(z - h)}$ $\\$ $\therefore \sqrt{v}\ \infty\ z$ $\\$ Second method : $\\$ We can directly get value of $v$ by $\\$ ` $hv =$ Energy $\\$ $\Rightarrow v = \dfrac{Energy (in\ kev)}{h}$ This we have to find out $\sqrt{v}$ and draw the same graph as above.

17   Use Moseley's law with $b - 1$ to find the frequency of the $K_0\ X$-ray of $La(Z - 57)$ if the frequency of the $K_a\ X$-ray of $Cu(Z - 29)$ is known to be $1\cdot88 \times 10^{18}\ Hz.$

##### Solution :

$b = 1$ $\\$ For $\infty\ a\ (57)$ $\\$ $\sqrt{v} = a\ (Z – b)$ $\\$ $\Rightarrow \sqrt{v} = a\ (57 – 1) = a \times 56 \ \ \ \ \ \ \ \ \ \ \ \ \ …(1)$ $\\$ For Cu$(29)$ $\\$ $\ \ \ \ \ \ \sqrt{1.88 \times 10^{78}} = a(29 - 1) = 28 a \ \ \ \ \ \ \ \ \ .....(2)$ $\\$ dividing $(1)$ and $(2)$ $\\$ $\sqrt{\dfrac{v}{1.88 \times 10^18}} = \dfrac{a \times 56}{a \times 28} = 2.$ $\\$ $\Rightarrow v = 1.88 \times 10^8(2)^2 = 4 \times 1.88 \times 10^18 = 7.52 \times 10^8\ Hz$

18   The $K_a$ and $K_p\ X$-rays of molybdenum have wavelengths $0\cdot71\ \dot{A}$ and $0\cdot63\ \dot{A}$ respectively. Find the wavelength of $L_a\ X$-ray of molybdenum.

##### Solution :

$K_{\alpha} = E_K – E_L \ \ \ \ \ \ \ \ \ \ \ \ \ ,,,(1)$ $\ \ \ \ \ \ \ \lambda K_{\alpha} = 0.71\ A°$ $\\$ $K_{\beta} = E_K – E_M \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,,,(2)$ $\ \ \ \ \ \ \ \ \ \ \ \lambda K_{\beta} = 0.63\ A°$ $\\$ $L_{\alpha} = E_L – E_M \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,,,(3)$ $\\$ Subtracting $(2)$ from $(1)$ $\\$ $K_{\alpha} – K_{\beta} = E_M – E_L = –L_{\alpha}$

or, $L_{\alpha} = K_{\beta} – K_{\alpha} = \dfrac{3 \times 10^8}{0.63 \times 10^{-10}} - \dfrac{3 \times 10^8}{0.71 \times 10^{-10}}$ $\\$ $= 4.761 \times 10^{18} – 4.225 \times 10^{18} = 0.536 \times 10^{18}\ Hz.$ $\\$ Again $\lambda = \dfrac{3 \times 10^8}{0.536 \times 10^{18}} = 5.6 \times 10^{–10} = 5.6\ A°.$

19   The wavelengths of $K_a$ and $L_a\ X$-rays of a material are $21\cdot3$ pm and $141$ pm respectively. Find the wavelength of $K_p\ X$-ray of the material.

##### Solution :

$E_1 = \dfrac{1242}{21.3 \times10^{-3}} = 58.309 \times 103\ ev$ $\\$ $E_2 = \dfrac{1242}{141 \times 10^{-3}}= 8.8085 \times 10^3\ ev$ $\\$ $E_3 = E_1 + E_2 \Rightarrow (58.309 + 8.809)\ ev = 67.118 \times 10^3\ ev$ $\\$ $\lambda = \dfrac{hc}{E_3} = \dfrac{1242}{67.118 \times 10^3} = 18.5 \times 10^{–3}\ nm = 18.5\ pm$.

20   The energy of a silver atom with a vacancy in $K$ shell is $25\cdot31\ keV,$ in $L$ shell is $3\cdot56\ keV$ and in $M$ shell is $0\cdot530\ keV$ higher than the energy of the atom with no vacancy. Find the frequency of $K_a,\ K_p$ and $L_a\ X$-rays of silver.

##### Solution :

$E_K = 25.31 KeV,\ E_L = 3.56\ KeV,\ E_M = 0.530\ KeV$ $\\$ $K_{\alpha} = E_K – K_L = hv$ $\\$ $\Rightarrow v = \dfrac{E_K - E_L}{h} = \dfrac{25.31 \times 0.53}{4.14 \times 10^{-15}} \times 10^3 = 5.25 \times 10^{15}\ Hz$ $\\$ $K_{\beta} = E_K – K_M = hv$ $\\$ $\Rightarrow v = \dfrac{E_K - E_M }{h} = \dfrac{25.31 \times 0.53}{4.14 \times 10^{-15}} \times 10^3 = 5.985 \times 10^{18}\ Hz.$ $\\$

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 $keV$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 69.5 \ \ \ \ \ \ \ 11.3 \ \ \ \ \ \ \ 2.3$

##### Solution :

Let for, $k$ series emission the potential required $= v$ $\\$ $\therefore$ Energy of electrons $= ev$ $\\$ This amount of energy $ev =$ energy of $L$ shell $\\$ The maximum potential difference that can be applied without emitting any electron is $11.3\ ev.$

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 electrons hitting the 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 ?

##### Solution :

$V = 40\ KV,\ i = 10\ mA$ $1$% of $T_{KE}$ (Total Kinetic Energy) $= X$ ray $\\$ $i = ne \ \ \ \ \ \ \ \ \ \ \ \ \$ or $n = \dfrac{10^{2}}{1.6 \times 10^{-19}} = 0.625 \times 10^{17}$ no.of electrons. $\\$ $KE$ of one electron $= eV = 1.6 \times 10^{–19} \times 40 \times 10^3 = 6.4 \times 10^{–15}\ J$ $\\$ $T_{KE} = 0.625 \times 6.4 \times 1017 \times 10^{–15} = 4 \times 10^2\ J.$ $\\$ a) Power emitted in $X$-ray $= 4 \times 10^2 \times (–1/100) = 4w$ $\\$ b) Heat produced in target per second $= 400 – 4 = 396\ J.$

Heat produced/sec $= 200\ w$ $\Rightarrow {neV}{t} = 200 \Rightarrow (ne/t)V = 200$ $\\$ $\Rightarrow i = 200 /V = 10\ mA.$

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.

##### Solution :

Heat produced/sec $= 200\ w$ $\\$ $\Rightarrow {neV}{t} = 200 \Rightarrow (ne/t)V = 200$ $\\$ $\Rightarrow i = 200 /V = 10\ mA.$

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 subsequently emitted. The emitted $X$-rays are analysed and the intensity is plotted against the wavelength (figure 44-E1). Assuming that only $K_{\alpha}$ intensities are detected, list the elements present in the sample from the plot. Use Moseley's equation. $$v = (25 \times 10^{14}\ Hz)\ (Z\ -\ 1)^2.$$

##### Solution :

Given : $v = (25 \times 10^{14}\ Hz)(Z – 1)^2$ $\\$ Or $C/\lambda = 25 \times 10^{14}\ (Z – 1)^2$ $\\$ a) $\dfrac{3 \times 10^8}{78.9 \times 10^{-12} \times 25 \times 10^{14}} = (Z - 1)^2$ $\\$ or, $(Z – 1)^2 = 0.001520 \times 10^6 = 1520$ $\\$ $\Rightarrow Z – 1 = 38.98\ or\ Z = 39.98 = 40.$ It is $(Zr)$ $\\$

b) $\dfrac{3 \times 10^8}{146 \times 10^{-12} \times 25 \times 10^{14}} = (Z - 1)^2$ $\\$ or, $(Z – 1)^2 = 0.0008219 \times 10^6$ $\Rightarrow Z – 1 = 28.669$ or\ $Z = 29.669 = 30$. It is $(Zn)$. $\\$ c) $\dfrac{3 \times 10^8}{158 \times 10^{-12} \times 25 \times 10^{14}} = (Z - 1)^2$ $\\$ or, $(Z – 1)^2 = 0.0007594 \times 10^6$ $\\$ $\Rightarrow Z – 1 = 27.5589$ or $Z = 28.5589 = 29.$ It is $(Cu)$. $\\$ d) $\dfrac{3 \times 10^8}{198 \times 10^{-12} \times 25 \times 10^{14}} = (Z - 1)^2$ $\\$ or, $(Z – 1)^2 = 0.000606 \times 10^6$ $\\$ $\Rightarrow Z – 1 = 24.6182$ or $Z = 25.6182 = 26$. It is $(Fe).$

25   A free atom of iron emits $K_a\ X$-rays of energy $6\cdot4\ keV$. Calculate the recoil kinetic energy of the atom. Mass of an iron atom = $9\cdot3 \times 10^{20}\ kg$.

##### Solution :

Here energy of photon $= E$ $\\$ $E = 6.4\ KeV = 6.4 \times 10^3\ ev$ $\\$ Momentum of Photon $= E/C = {6.4 \times 10^3}{3 \times 10^8} = 3.41 \times 10^{–24}$ m/sec. $\\$ According to collision theory of momentum of photon = momentum of atom $\\$ $\therefore$ Momentum of Atom $= P = 3.41 \times 10^{–24}$ m/sec $\\$ $\therefore$ Recoil $K.E.$ of atom $= P^2 / 2m$ $\\$ $\Rightarrow \dfrac{(3.41 \times 10^{-24})^2\ eV}{(2)(9.3 \times 10^{-16} \times 1.6 \times 10^{-19})} = 3.9\ eV\ [1$ Joule $= 1.6 \times 10{–19}\ ev]$

26   The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength $(1/\lambda.)$ 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 $(I/\lambda)$ of the $X$-ray emitted. Show that the slopes of the lines in the two cases are equal and find its value.

##### Solution :

$V_0 \rightarrow$ Stopping Potential, $\lambda \rightarrow$ Wavelength, $eV_0 = hv – hv_0$ $\\$ $eV_0 = \dfrac{hc}{\lambda} \Rightarrow V_0\lambda = \dfrac{hc}{e}$ $\\$ $V\ \rightarrow$ Potential difference across X-ray tube, $\lambda \rightarrow$ Cut of wavelength $\\$ $\lambda = \dfrac{hc}{eV} \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ V\lambda = \dfrac{hc}{e}$ $\\$ Slopes are same i.e. $V_0\lambda = V\lambda$ $\\$ $\dfrac{hc}{e} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19}} = 1.242 \times 10{–6}\ Vm$

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 $0\cdot1\ mm$?

##### Solution :

$\lambda = 10\ pm = 100 \times 10^{–12}\ m$ $\\$ $D = 40\ cm = 40 \times 10^{–2}\ m$ $\\$ $\beta = 0.1\ mm = 0.1 \times 10^{–3}\ m$ $\\$ $\beta = \dfrac{\lambda{D}}{d}$ $\\$ $\Rightarrow d = \dfrac{\lambda{D}}{\beta} = \dfrac{100 \times 10^{-12} \times 40 \times 10^{-2}}{10^{-3} \times 0.1} = 4 \times 10^{–7}\ m.$