Question

Find the de Broglie wavelength associated with a thermal neutron of mass \(m\) at absolute temperature \(T\). \[ k=\text{Boltzmann constant}, \qquad h=\text{Planck's constant} \]

• A. \[ \frac{h}{\sqrt{3mkT}} \]
• B. \[ \frac{h}{\sqrt{2mkT}} \]
• C. \[ \frac{h}{3mkT} \]
• D. \[ \frac{h}{\sqrt{mkT}} \]

Hint:

For a thermal particle, \[ \frac{p^2}{2m} = \frac32 kT. \] Then use \[ \lambda=\frac{h}{p}. \]

Answer 1

The Correct Option is A

Concept: For a thermal particle, \[ \text{Average kinetic energy} = \frac32 kT. \] Also, \[ K=\frac{p^2}{2m}. \] The de Broglie wavelength is \[ \lambda=\frac{h}{p}. \]

Step 1: Relate momentum and temperature. \[ \frac{p^2}{2m} = \frac32 kT. \] Multiplying by \(2m\), \[ p^2 = 3mkT. \] Hence, \[ p=\sqrt{3mkT}. \]

Step 2: Apply the de Broglie relation. \[ \lambda = \frac{h}{p} \] \[ = \frac{h}{\sqrt{3mkT}}. \] \[\begin{aligned} \boxed{ \lambda= \frac{h}{\sqrt{3mkT}} } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.