Question

Electrons of mass m with de-Broglie wavelength $ \lambda $ fall on the target in an X-rays tube. The cutoff wavelength $ {\lambda _0} $ of the emitted X-rays is
A) $ {\lambda _0} = \lambda $
B) $ {\lambda _0} = \dfrac{{2mc{\lambda ^2}}}{h} $
C) $ {\lambda _0} = \dfrac{{2h}}{{mc}} $
D) $ {\lambda _0} = \dfrac{{2{m^2}{c^2}{\lambda ^3}}}{{{h^2}}} $

Answer 1

Hint When electrons fall on the target in an X-ray tube it experiences various interactions however the cut-off wavelength is associated with the maximum energy of the emitted x-rays. Use de Broglie's equation to calculate the energy of the incident electrons and then calculate the wavelength of x-rays emitted due to the energy equal to the energy of the incident electrons.

Formula used:
-De-Broglie equation: $ \lambda = \dfrac{h}{p} $ where $ \lambda $ is the wavelength of the electrons and $ p $ is their momentum
-Energy of x-rays: $ E = \dfrac{{hc}}{\lambda } $ where $ h $ is the Planck’s constant, $ c $ is the speed of light, $ \lambda $ is the wavelength of the emitted X-rays.

Complete step by step answer
When electrons interact with a target, they excite other electrons and cause electronic transitions inside the atom which releases x-rays. The energy of the emitted x-rays does not have a single value but rather a spectrum. However, the maximum energy that can be emitted is equal to the energy carried by the incident electron because of the law of conservation of energy.
So the energy associated with electrons with de-Broglie wavelength $ \lambda $ can be calculated using the de-Broglie equation as:
 $\Rightarrow \lambda = \dfrac{h}{p} $
 $\Rightarrow \lambda = \dfrac{h}{{\sqrt {2mE} }} $ where $ m $ is the mass of the electron and $ E $ is its energy.
Since we know that the cut-off wavelength of emitted x-rays corresponds to the incoming energy of electrons, we can write:
 $\Rightarrow E = \dfrac{{hc}}{{{\lambda _0}}} $
Substituting the value of incoming energy of electrons in de Broglie's equation, we get:
 $\Rightarrow \lambda = \dfrac{h}{{\sqrt {2m\dfrac{{hc}}{{{\lambda _0}}}} }} $
Solving for $ {\lambda _0} $ , we get
$ {\lambda _0} = \dfrac{{2mc{\lambda ^2}}}{h} $ which corresponds to option (B).

Note
The trick in solving this question is realizing that the emitted x-rays will have a cut-off wavelength such that their energy will be equal to the energy of incoming electrons. While option (A) is tempting, we must remember that the energy of EM waves like x-rays are different from the energy of particles like electrons.