Question

A beam of electrons is accelerated by a potential difference V is made to strike on a metal target to produce X-ray. For which of the following value of V, the resulting x-ray have the lowest possible wavelength:
(a) 10KV
(b)20KV
(c) 30KV
(d) 40KV

Answer 1

Hint: Higher the energy of emitted radiation shorter the wavelength. So, use energy conservation and use the least wavelength possible for X-ray from the spectrum. To get upper bound for voltage possible.

Formula Used:
The energy of any radiation is given by $E = hv$ …… (A)
Where E is the energy of radiation, $h$ is the Planck’s constant and $v$ is the frequency of the radiation.
In terms of wavelength of radiation, Energy of radiation is given by: $E = \dfrac{{hc}}{\lambda }$ …… (B)
Where, $E$ is the energy of radiation, $h$ Planck’s constant, $c$ is the speed of light or radiation, and $\lambda $ is the wavelength of light emitted.
The maximum energy of ejected radiation= Total energy incident onto the metal plate.

Complete step by step answer:
Given,
Acceleration voltage =$V$
Charge of electron =$e$
Step 1:
Total energy is conserved in the entire process:
Therefore, ${\left. {T.E} \right|_{initial}} = {\left. {T.E} \right|_{final}}$ …… (1)
Total initial energy of the system,${\left. {T.E} \right|_{initial}} = eV$ …… (2)
Total final energy of the system,
${\left. {T.E} \right|_{final}} = \dfrac{{hc}}{{{\lambda _{radiation}}}} + K.{E_{{\text{of particle emitted}}}}$ …… (3)
Step 2:
Let’s say kinetic energy of emitted particle is 0 (just for simplicity otherwise, in real world no particle emerges out with 0 K.E)
${\left. {T.E} \right|_{final}} = \dfrac{{hc}}{{{\lambda _{radiation}}}}$ …… (4)
Then, we can also say,
${\left. { \Rightarrow T.E} \right|_{final}} \propto \dfrac{1}{{{\lambda _{radiation}}}}$ ………… (5)
From equation (1) and (4) we can say,
${\left. {T.E} \right|_{final}} = \dfrac{{hc}}{{{\lambda _{radiation}}}}$
Step 3:
Using equation (2) and equation (4) we can say-
$ \Rightarrow eV = \dfrac{{hc}}{{{\lambda _{radiation}}}}$
$ \Rightarrow V = \dfrac{{hc}}{{e{\lambda _{radiation}}}}$
$ \Rightarrow V \propto \dfrac{1}{{{\lambda _{radiation}}}}$ …… (6) ( hence, V inversely varies with wavelength)
Step 4:
We know range of X rays wavelength lies between: 10 nm to 10pm (i.e ${10^{ - 9}}m{\text{ to }}{10^{ - 11}}m$ in general from EM wave spectrum)
So, putting the value of planks constant and charge on the electron we get an upper bound on values of V by putting the lowest possible value of wavelength in equation (6) to fall in the above range.
\[ \Rightarrow V < \dfrac{{(6.6 \times {{10}^{ - 34}}J.s)(3 \times {{10}^8}m{s^{ - 1}})}}{{(1.6 \times {{10}^{19}}C){\lambda _{radiation}}}}\]
\[ \Rightarrow V < \dfrac{{(12.375 \times {{10}^{ - 7}})Volt.m}}{{{\lambda _{radiation}}}}\]
\[ \Rightarrow V < \dfrac{{(12.375 \times {{10}^{ - 7}})Volt.m}}{{{{10}^{ - 11}}m}}\]
\[ \Rightarrow V < 12.375 \times {10^4})Volt\]
\[ \Rightarrow V < 123.75KV\] (max limit) …… (7)

Therefore, up to $123KV$ X-rays can be produced by the acceleration of electrons through potential. So, using equation (7) we can say, 40KV would give the least wavelength among all.

Therefore, the correct option is (D) 40KV.

Note:
Here, we used the K.E. term to be $0$ because we have first checked whether there would be any X-ray coming out corresponding to the given voltage range. Similarly, we can find lower bound as well for X-ray production in the above case.