If the kinetic energy of sa particle is reduced to half .Eebroglie wave length becomes.

Options

A. 2 times
B. `(1)/(sqrt2)` times
C. 4 times
D. `sqrt(2)` times

Answer

D

Explanation

The de-Brogile wavelength is inversely proportional to the square root of kinetic energy.
Hence, `(lambda_1)/(lambda_2)=sqrt(((KE)_2)/((KE)_1))`
Given that, `(KE)_(2)=(1)/(2)(KE)_(1)`
Then, `(lambda_1)/(lambda_2) "or",lambda_(2)=sqrt(2)lambda_(1)`
Thus, de-Brogile wavelength becomes `sqrt(2)` times to the original wavelength.