Solveeit Logo
Login

Question

Question: An electron with initial kinetic energy of 100 eV is accelerated through a potential difference of 5...

An electron with initial kinetic energy of 100 eV is accelerated through a potential difference of 50 V. Then find the de-Broglie wavelength of electron becomes
(A) 1A∘1\text{A}{}^\circ
(B) 1.5A∘\sqrt{1.5}\text{A}{}^\circ
(C) 3A∘\sqrt{3}\text{A}{}^\circ
(D) 12⋅27A∘12\cdot 27\text{A}{}^\circ

Explanation

Solution

The wave associated with moving particles is called matter wave or de-Broglie wave whose wavelength is called de-Broglie wavelength.
λ=hmv\lambda =\dfrac{h}{mv}
Where h is planck's constant, m is mass of particle and v is velocity of particle.
The kinetic energy is given by
v=2Km λ=hm2k/m=h2km \begin{aligned} & v=\sqrt{\dfrac{2K}{m}} \\\ & \lambda =\dfrac{h}{m\sqrt{2k/m}}=\dfrac{h}{\sqrt{2km}} \\\ \end{aligned}
Using the above formula, we will get the required result.

Complete step by step solution:
We have given, an electron having kinetic energy = 100 eV
An electron is accelerated through a potential difference, So additional kinetic energy is gains =e(50V)=50ev=\text{e}\left( 50\text{V} \right)=50\text{ev}
Total kinetic energy is equal to =50=50 ev +100+100 ev =150=150 ev
=150×1⋅6×10−19=150\times 1\cdot 6\times {{10}^{-19}} J
The de-Broglie wavelength can be calculated from the following formula.
λ=hmv λ=h2mk \begin{aligned} & \lambda =\dfrac{h}{mv} \\\ & \lambda =\dfrac{h}{\sqrt{2mk}} \\\ \end{aligned} --- (1)
Mass of electron = 9.11×10−31kg9.11\times {{10}^{-31}}kg
Planck's constant = 6.63×10−346.63\times {{10}^{-34}} Js
Put all the values in eq. (1)
λ=6.63×10−342×9.11×10−31×150×1.6×10−19 λ=6.63×10−344372.8×10−50 λ=6.63×10−3466.13×10−25 λ=6.63×10−346.613×10−24 λ=10−10m \begin{aligned} & \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{2\times 9.11\times {{10}^{-31}}\times 150\times 1.6\times {{10}^{-19}}}} \\\ & \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{4372.8\times {{10}^{-50}}}} \\\ & \lambda =\dfrac{6.63\times {{10}^{-34}}}{66.13\times {{10}^{-25}}} \\\ & \lambda =\dfrac{6.63\times {{10}^{-34}}}{6.613\times {{10}^{-24}}} \\\ & \lambda ={{10}^{-10}}m \\\ \end{aligned}
λ=1A∘\lambda =1\text{A}{}^\circ
Hence, the de-Broglie wavelength is 1A∘1\text{A}{}^\circ .
So, the correct option is “A”.

Note:
Significance of de-Broglie equation.
De-Broglie proposed that as light exhibits both wave-like and particle-like properties, matter exhibits wave-like and particle-like properties. This nature was described as dual behavior of matter.
Another method.
Use direct formula, λ=12.27v\lambda =\dfrac{12.27}{\sqrt{v}}
V can be calculated as 12mv2=eV\dfrac{1}{2}m{{v}^{2}}=eV , Put the values in the above formula, we will get the same result.