Question

If the kinetic energy of free electron is made double; the new de-Broglie wavelength will be __________ times that of initial wavelength.

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

We know that
de-Broglie wavelength $\lambda =\frac{h}{p}$
$\lambda =\frac{h}{mv}$
If $K$ be the kinetic energy of the electron, then
$K=\frac{1}{2}m{v}^2$
$v=\sqrt{2K/m}$
So, de-Broglie wavelength is
$\lambda =\frac{h}{\sqrt{2mK}}$
$\lambda \propto \frac{1}{\sqrt{K}}$
$\therefore {\lambda }_1=\frac{1}{\sqrt{K_1}}$ ....(i)
${\lambda }_2=\frac{1}{\sqrt{2K_1}}$ ....(ii)
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{1/\sqrt{2K_1}}{1/\sqrt{K_1}}$
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{1}{\sqrt{2}}$
So, the new de-Broglie wavelength will be $\frac{1}{\sqrt{2}}$ times that of initial wavelength.