Question

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

Answer 1

In order to this question, to find the cut-off wavelength, first we will assume the kinetic energy which is used to find the momentum and as it is related to the de-Broglie’s equation then we will find the kinetic energy in terms of wavelength. After that we will also find kinetic energy in terms of cutoff wavelength using kinetic energy of photons.At last we will equate both kinetic energy equations to find the required relation.

Complete step by step answer:
Let the Kinetic energy of electrons be $KE$. As we all know that de-Broglie relates with the momentum, so we have to know the equation of momentum.The relation between kinetic energy and momentum is:
$K.E = \dfrac{{{p^2}}}{{2m}}$
where, $p$ is the momentum and $m$ is the mass of electrons.
By applying de-Broglie wavelength equation-related to electron is:
$\because \lambda = \dfrac{h}{p} \\\ \Rightarrow \lambda = \dfrac{h}{{\sqrt {2mKE} }} \\\ \Rightarrow KE = \dfrac{{{h^2}}}{{2m\lambda }} \\\$---(eq1)
Now, the relation between cut-off wavelength of emitted X-rays is related to kinetic energy of incident electron as
$\dfrac{{hc}}{{{\lambda _0}}} = K.E$ ….(eq2)
Now, from eq(1) and (2):-
$\dfrac{{{h^2}}}{{2m\lambda }} = \dfrac{{hc}}{{{\lambda _0}}} \\\ \therefore {\lambda _0} = \dfrac{{2mc{\lambda ^2}}}{h}$

Hence, the correct answer is option A.

Note: According to the de Broglie equation, matter can behave like light and radiation, which are both waves and particles. The equation goes on to say that a beam of electrons can be diffracted in the same way as a beam of light may. In a nutshell, the de Broglie equation clarifies the concept of matter having a wavelength.