Question

De Broglie wavelength $\lambda$ associated with neutrons is related with absolute temperature T as:

• A. $\lambda \propto T$
• B. $\lambda \propto \frac{1}{T}$
• C. $\lambda \propto \frac{1}{\sqrt{T}}$
• D. $\lambda \propto T^2$

Answer 1

De broglie wavelength $\lambda =\frac{h}{mv}$

where $v$ is the velocity of the particle.

Let $E$ be the kinetic energy of the neutrons.

Thus $E=\frac{1}{2}m{v}^2$

$⟹v=\sqrt{\frac{2E}{m}}$

Therefore $\lambda =\frac{h}{\sqrt{2mE}}$

Since energy of neutrons $E\propto T$

$\lambda \propto \frac{1}{\sqrt{T}}$