Question

If the kinetic energy of an electron is increased to 16 times the initial value, how many times would the new de Broglie wavelength be?

• A. 4
• B. 0.25
• C. 16
• D. 0.5

Answer 1

The correct option is B 0.25

de Broglie's wavelength in terms of kinetic energy can be given as: $\lambda =\frac{h}{\sqrt{2K.E.m}}$

Let $K.E_i\text{and}K.E_f$ be the initial and final kinetic energies while ${\lambda }_i,{\lambda }_f$ are the de Broglie's wavelengths for respective kintetic energies.

It is clear that:
$\frac{{\lambda }_f}{{\lambda }_i}=\frac{\sqrt{K.E_i}}{\sqrt{K.E_f}}$

Given that:
$K.E_f=16\times K.E_i$

So,
$\frac{{\lambda }_f}{{\lambda }_i}=\frac{\sqrt{K.E_i}}{\sqrt{16\times K.E_i}}=\frac{1}{4}\therefore {\lambda }_f=\frac{1}{4}{\lambda }_i$