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.
\[m{{c}^{2}}=\dfrac{hc}{\lambda }=ev\]
\[\Rightarrow m{{c}^{2}}=ev\]
\[\Rightarrow m=\dfrac{ev}{{{c}^{2}}}\]
\[e=1.6\times {{10}^{-19}}c\]
\[\Rightarrow m=\dfrac{1.6\times {{10}^{-19}}\times 1.8\times {{10}^{3}}}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}\]
\[\Rightarrow m=3.2\times {{10}^{-32}}\text{ kg}\]
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.
\[m{{c}^{2}}=\dfrac{hc}{\lambda }=eV\]