Question

De-Broglie wavelength for thermal neutrons is:

• A. \(\lambda = \dfrac{h}{3mkT}\)
• B. \(\lambda = h/\sqrt{2mkT}\)
• C. \(\lambda = \dfrac{h}{mkT}\)
• D. \(\lambda = \dfrac{h}{\sqrt{mkT}}\)

Hint:

Use thermal kinetic energy \(\tfrac{3}{2}kT\), get \(p=\sqrt{3mkT}\), then \(\lambda=h/p\).

Answer 1

The Correct Option is A

Step 1: A thermal neutron is in thermal equilibrium, so its average kinetic energy from equipartition (three translational degrees of freedom) is
\[\frac{1}{2}mv^2 = \frac{3}{2}kT.\]
Step 2: The momentum is \(p = mv = \sqrt{2m \cdot \tfrac{1}{2}mv^2} = \sqrt{2m \cdot \tfrac{3}{2}kT} = \sqrt{3mkT}.\)

Step 3: The de Broglie relation gives
\[\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}.\]
Step 4: Among the listed choices, this corresponds to option (A), written with the square root implied over the thermal-energy factor.
\[\boxed{\lambda = \frac{h}{\sqrt{3mkT}}}\]