If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will be _____

Given Data:

The energy of a particle is reduced to one fourth.

Formula Used:

$\frac{1}{2}m{v}^2=E$

Detailed Solution:

$\frac{1}{2}m{v}^2=E$

$mv=\sqrt{2mE} \& \lambda =\frac{h}{mv}$

$\lambda \prime =\frac{h}{\sqrt{2mE\text{'}}}$
$\lambda \prime =\frac{h}{\sqrt{2m{\displaystyle \frac{E}{4}}}}$
$\frac{\lambda \prime }{\lambda }=2$
$\lambda \prime =2\lambda$

Hence, the correct answer is 100%.