The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m$, is :

Question

The variation of the stopping potential $ V_0 $ with the frequency $ \nu $ of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for which metal?
variation of the stopping potential

Answer 1

To find the de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $ T $ (Kelvin), we can use the de-Broglie wavelength formula, which relates the wavelength $ \lambda $ to the momentum $ p $ of a particle:

$\lambda = \frac{h}{p}$

where $ h $ is Planck's constant.

For a neutron in thermal equilibrium, we can approximate its momentum using the equipartition theorem. The kinetic energy $ E $ of the neutron is given by:

E = \frac{3}{2}kT

The momentum $ p $ can be related to the kinetic energy by the equation:

E = \frac{p^2}{2m}

Equating the two expressions for kinetic energy, we have:

\frac{p^2}{2m} = \frac{3}{2}kT

Solving for $ p $, we find:

p = \sqrt{3m kT}

Substitute this expression for $ p $ into the de-Broglie wavelength formula:

\lambda = \frac{h}{\sqrt{3m kT}}

This corresponds to the option:

$\frac{h}{\sqrt{3m kT}}$

Thus, the correct answer is $\frac{h}{\sqrt{3m kT}}$.

Answer 2