Question

The de Broglie wavelength of a neutron when its kinetic energy is $K$, is $\lambda$. What will be its wavelength when its kinetic energy is $4K$?

• A. $\frac{\lambda }{4}$
• B. $\frac{\lambda }{2}$
• C. $2\lambda$
• D. $4\lambda$

Answer 1

The correct option is B $\frac{\lambda }{2}$

Kinetic energy of neutron=$\frac{1}{2}m{v}^2=K$

$⟹v=\sqrt{\frac{2K}{m}}$

De-broglie wavelength of neturon=$\frac{h}{p}$

$=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}\propto \frac{1}{\sqrt{K}}$

Thus when kinetic energy becomes $4K$, the de-broglie wavelength becomes $\frac{\lambda }{2}$.