The cut- off wavelength when a potential difference of $25\,kV$ is applied to an $X$ - ray tube, is:
A) $0.248 A° $
B) $0.496 A° $
C) $0.124 A° $
D) $0.620 A° $

Hint: Use the formula of the cut of wavelength of the $X$ - ray tube given below and substitute all the constants in the formula and the potential difference in the formula and calculate it to find the cut off wavelength of the $X$ - ray tube.

Formula used:
The formula of the cut off wavelength is given by
$\lambda = \dfrac{{hc}}{{Ve}}$
Where $\lambda $ is the cut off wavelength, $h$ is the Planck constant, $c$ is the speed of the light and $V$ is the accelerating potential and $e$ is the charge of the electron.

Complete step by step solution:
It is given that the
The potential difference applied on the $X$ - ray tube, $V = 25\,KV = 25000\,V$
Using the formula of the cut off wavelength,
$\lambda = \dfrac{{hc}}{{Ve}}$

The value of the Planck’s constant that is substituted in the above cut off wavelength calculation is $6.64 \times {10^{ - 34}}$ . The charge of the electron that is substituted as $1.6 \times {10^{ - 19}}$. The speed of the light is substituted as $3 \times {10^8}\,m{s^{ - 1}}$. All these values are also substituted as the constant $12400$ . Substituting the known values in the above formula,

$\lambda = \dfrac{{6.64 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{25 \times {{10}^3} \times 1.6 \times {{10}^{ - 19}}}}$
By performing the simplification in the above step,
$\lambda = \dfrac{{19.92 \times {{10}^{ - 26}}}}{{40 \times {{10}^{ - 16}}}}$
By further simplification of the above equation,
$\lambda = 0.496 \times {10^{10}}$
Hence the above answer is also written as $0.496 A° $.

Thus the option (B) is correct.

Note: The $X$ - ray tube emits the $X$ - rays at the minimum wavelength called as the cut off wavelength. The intensity of the $X$ - ray increases with the increases with the accelerating potential and the cut off wavelength decreases with the increase in the accelerating potential.