Question

The voltage applied to an x-ray to be is 18 KV. The maximum mass of photon emitted by the X – ray tube will be:
A.$$2\times {10}^{-13}\text{ kg}$$
B. $$\text{3}\text{.2}\times {10}^{-32}\text{ kg}$$
C. $$\text{6}\text{.4}\times {10}^{-32}\text{ kg}$$
D. $$\text{9}\text{.1}\times {10}^{-31}\text{ kg}$$

Answer 1

Hint: Basically, in X-ray the voltage applied between cathode and anode. In this question we equate the energy of Photon with the maximum energy of an electron. The maximum energy of a photon is $m{c}^2$.

Complete step by step answer:
X-rays were produced by Wilhelm Roentgen in 1895 while testing whether cathode rays pass through glass. When the voltage is applied through the x-ray tube or between cathode and anode then maximum energy of the photon remains equal to the maximum energy of the photon.

Given applied voltage $$=18KV=1\cdot 8\times {10}^3\text{ volt}$$
As we know the energy of photon$$=m{c}^2$$
Where, m = mass of photon
c = speed of light
Also the maximum energy of the photon = maximum energy of electron.
$$m{c}^2={\displaystyle \frac{hc}{\lambda }}=ev$$
$$\Rightarrow m{c}^2=ev$$
$$\Rightarrow m={\displaystyle \frac{ev}{c^2}}$$
$$e=1.6\times {10}^{-19}c$$
$$\Rightarrow m={\displaystyle \frac{1.6\times {10}^{-19}\times 1.8\times {10}^3}{{(3\times {10}^8)}^2}}$$
$$\Rightarrow m=3.2\times {10}^{-32}\text{ kg}$$
Hence, the maximum mass of photon is $$3.2\times {10}^{-32}\text{ kg}$$

Therefore, the correct choice is (B) $$3.2\times {10}^{-32}\text{ kg}$$

Note:
When a voltage is applied to an X – ray tube (b/w cathode and anode) then the maximum energy of the emitted photon remains equal to the energy of the electron.
$$m{c}^2={\displaystyle \frac{hc}{\lambda }}=eV$$