Question

If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will be:
A. 41%
B. 141%
C. 100%
D. 71%

Answer 1

Hint: Firstly we have to find some relationship that relates the given quantities such that we could find how decrease in one would affect the other. Then we could recall how to find the percentage increase in a quantity by knowing its initial and final values and hence find the answer.
Formula used:
Momentum,
$p=\sqrt{2mK.E}$
De Broglie wavelength,
$\lambda =\dfrac{h}{p}$

Complete answer:
In the question, we are given that the energy of a particle is reduced to one fourth of its initial value. We are supposed to find the percentage increase in its de Broglie wavelength.
First of all we have to find a relation that relates the kinetic energy of a particle with its de Broglie wavelength.
We know that the momentum of a particle is given by,
$p=\sqrt{2mK.E}$ ………………………………… (1)
We also know that the de Broglie wavelength is given by,
$\lambda =\dfrac{h}{p}$ ………………………………………. (2)
Combining (1) and (2) we get,
$\lambda =\dfrac{h}{\sqrt{2mK.E}}$ ………………………………………. (3)
Let this be the initial de Broglie wavelength. Now when the energy is reduced to fourth of its initial value, the de Broglie wavelength would be,
$\lambda '=\dfrac{h}{\sqrt{2m\left( \dfrac{K.E}{4} \right)}}$
$\Rightarrow \lambda '=\dfrac{2h}{\sqrt{2mK.E}}=2\lambda $
Now, for finding the percentage increase in de Broglie wavelength, we have,
$\dfrac{\lambda '-\lambda }{\lambda }\times 100=\dfrac{2\lambda -\lambda }{\lambda }\times 100=100$
Therefore, we found the percentage increase to be 100% when the energy of the particle is reduced to one fourth.

Option C is correct.

Note:
For any physical quantity the percentage increase is found in the same way.
$\dfrac{final-initial}{initial}\times 100$
All the other relationships used in the above solution are very basic and should be known by heart in order to solve these kinds of problems.