Courses for Kids
Free study material
Offline Centres
Store Icon

In an X-ray tube, electrons bombarding the target produce an X-ray of minimum wavelength $1\mathop {\text{A}}\limits^o $. The energy of the bombarding electron will be:
(A) 100 eV
(B) 14375 eV
(C) 12000 eV
(D) 12375 eV

Last updated date: 19th Jul 2024
Total views: 62.4k
Views today: 0.62k
62.4k+ views
HintHere the energy of the electron bombarding on the target will be given by the energy of the X-ray photons which are released using the wavelength of the X-ray given in the question from the formula $E = \dfrac{{hc}}{\lambda }$

Formula Used
In this solution we will be using the following formula,
$E = h\upsilon $
where $h$ is the Planck’s constant
$E$ is the energy and $\upsilon $ is the frequency of the wave
and $\upsilon = \dfrac{c}{\lambda }$ where $\lambda $ is the wavelength of the wave and $c$ is the velocity of light.

Complete step-by-step answer:
When an electron bombards the target, the kinetic energy of the electron transfers in the interaction and a X-ray with the highest possible energy is released.
So the energy of the electron will be the energy of the X-ray. Now from the energy of the X-ray, we can find its frequency by the formula
$E = h\upsilon $ where $\upsilon $ is the frequency of the wave
The frequency can be written as, $\upsilon = \dfrac{c}{\lambda }$
So substituting we get,
$E = \dfrac{{hc}}{\lambda }$
From the question we have, the wavelength as $\lambda = 1\mathop {\text{A}}\limits^o = 1 \times {10^{ - 10}}m$
and the value of the speed of light and the Planck’s constant are,
$c = 3 \times {10^8}m/s$ and $h = 6.6 \times {10^{ - 34}}{m^2}kg/s$
So substituting all the values we get
$E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1 \times {{10}^{ - 10}}}}$
On doing the calculation we get,
$E = 1.98 \times {10^{ - 15}}J$
But the options are given in electron-volt. So to convert, we divide the value obtained by a factor of $1.6 \times {10^{ - 19}}$. Hence we get,
$E = \dfrac{{1.98 \times {{10}^{ - 15}}}}{{1.6 \times {{10}^{ - 19}}}}eV$
On doing the division we get,
$E = 12375eV$

So the correct answer is option D.

Note: In an X-ray tube, current is passed through the tungsten filament which enables it to get heated up and releases electrons by thermionic emission. These electrons bombard the target which results in conversion of energy into heat and X-ray photons.