Question

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) $1\text{A}{}^{\circ }$
(B) $\sqrt{1.5}\text{A}{}^{\circ }$
(C) $\sqrt{3}\text{A}{}^{\circ }$
(D) $12\cdot 27\text{A}{}^{\circ }$

Answer 1

Hint: The wave associated with moving particles is called matter wave or de-Broglie wave whose wavelength is called de-Broglie wavelength.
 $\lambda ={\displaystyle \frac{h}{mv}}$
Where h is planck's constant, m is mass of particle and v is velocity of particle.
The kinetic energy is given by
 $\begin{array}{rl}& v=\sqrt{{\displaystyle \frac{2K}{m}}}\\ & \lambda ={\displaystyle \frac{h}{m\sqrt{2k/m}}}={\displaystyle \frac{h}{\sqrt{2km}}}\end{array}$
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 $=\text{e}\left(50\text{V}\right)=50\text{ev}$
Total kinetic energy is equal to $=50$ ev $+100$ ev $=150$ ev
 $=150\times 1\cdot 6\times {10}^{-19}$ J
The de-Broglie wavelength can be calculated from the following formula.
 $\begin{array}{rl}& \lambda ={\displaystyle \frac{h}{mv}}\\ & \lambda ={\displaystyle \frac{h}{\sqrt{2mk}}}\end{array}$ --- (1)
Mass of electron = $9.11\times {10}^{-31}kg$
Planck's constant = $6.63\times {10}^{-34}$ Js
Put all the values in eq. (1)
 $\begin{array}{rl}& \lambda ={\displaystyle \frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 150\times 1.6\times {10}^{-19}}}}\\ & \lambda ={\displaystyle \frac{6.63\times {10}^{-34}}{\sqrt{4372.8\times {10}^{-50}}}}\\ & \lambda ={\displaystyle \frac{6.63\times {10}^{-34}}{66.13\times {10}^{-25}}}\\ & \lambda ={\displaystyle \frac{6.63\times {10}^{-34}}{6.613\times {10}^{-24}}}\\ & \lambda ={10}^{-10}m\end{array}$
 $\lambda =1\text{A}{}^{\circ }$
Hence, the de-Broglie wavelength is $1\text{A}{}^{\circ }$ .
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, $\lambda ={\displaystyle \frac{12.27}{\sqrt{v}}}$
V can be calculated as ${\displaystyle \frac{1}{2}}m{v}^2=eV$ , Put the values in the above formula, we will get the same result.