Q. The wavelength of de-Broglie wave associated with a thermal neutron of mass $m$ at absolute temperature $T$ is given by (Here, $k$ is the Boltzmann constant)
A) $\frac{h}{\sqrt{2mkT}}$
B) $\frac{h}{\sqrt{mkT}}$
C) $\frac{h}{\sqrt{3kmT}}$
D) $\frac{h}{2\sqrt{mkT}}$
Explanation
(c) KE of thermal neutron, $\frac{1}{2}mv^{2} = \frac{3}{2}kT$
or $mv = \sqrt{3kmT}$; So, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3kmT}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.