Question

What is the ratio of the de-Broglie wavelength for electrons accelerated through $200$ volts and $50$ volts?
A. $1:2$
B. $2:1$
C. $3:10$
D. $10:3$

Answer 1

In order to solve this question we need to understand de-Broglie wavelength and matter waves. So de-Broglie stated that there is a wave associated with every matter but this wave is more prominent at microscopic level or quantum level. So the wave associated with it is known as de-Broglie wavelength. These waves have wavelengths which are very less at classical level but at quantum level these wave’s wavelength is comparable to quantum size objects hence these waves interact with each other and produce many phenomena.

Complete step by step answer:
Let the De-Broglie wavelength be $\lambda$ so according to de broglie matter wave’s wavelength can be defined as,
$\lambda = \dfrac{h}{p}$
Considering non relativistic case, kinetic energy of particle is,
$K = \dfrac{1}{2}m{v^2}$
Here “m” is the mass of a particle, “v” is speed.
Let the momentum of particle be, $p$ so from definition we get, $p = mv$
velocity is, $v = \dfrac{p}{m}$
Using value of velocity in kinetic energy expression we get,
$K = \dfrac{1}{2}m{(\dfrac{p}{m})^2}$
$\Rightarrow K = \dfrac{{{p^2}}}{{2m}}$
So momentum is, $p = \sqrt {2mK}$

Putting expression of momentum in wavelength we get,
$\lambda = \dfrac{h}{{\sqrt {2mK} }}$
Let the particle be accelerated in potential difference “V” we get, $K = eV$.Putting value of “K” in expression of De-Broglie we get,
$\lambda = \dfrac{h}{{\sqrt {2meV} }}$
For potential difference ${V_1} = 200V$
So, ${\lambda _1} = \dfrac{h}{{\sqrt {2me{V_1}} }} \to (i)$
For potential difference ${V_2} = 50V$
So, ${\lambda _2} = \dfrac{h}{{\sqrt {2me{V_2}} }} \to (ii)$
So for ratio, dividing equation (i) by (ii) we get, $\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{{V_2}}}{{{V_1}}}}$
Putting values we get,
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{200}}{{50}}}$
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt 4$
$\therefore \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{2}{1}$

So correct option is B.

Note: It should be remembered that, here we have neglected the relativistic speed of electrons for the sake of simplicity in solving questions. Matter waves or De-Broglie waves are responsible for producing luminescence on screen using electrons as electrons behave as matter waves according to De-Broglie.