Question

Energy of a particle at absolute temperature \( T \) is of the order of \( kT \). Calculate the wavelength of thermal neutrons at 27°C. Find the energy of a photon having the same wavelength. Here, \( k \) is the Boltzmann constant.

Hint:

The wavelength of thermal particles is related to their temperature, and de Broglie wavelength can be applied even for non-light particles like neutrons.

Answer 1

- Step 1: Find the thermal energy at 27°C. The average thermal energy of a particle is given by \( E \sim kT \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. - Temperature in Kelvin is: \[ T = 27 + 273 = 300 \, \text{K}. \] The Boltzmann constant \( k = 1.38 \times 10^{-23} \, \text{J/K} \), so the energy of the particle is: \[ E = kT = (1.38 \times 10^{-23}) \times 300 = 4.14 \times 10^{-21} \, \text{J}. \] - Step 2: Calculate the de Broglie wavelength of a neutron. The wavelength of the neutron is given by de Broglie's relation: \[ \lambda = \frac{h}{p}, \] where \( p = mv \) is the momentum and \( h \) is Planck's constant. The kinetic energy of the thermal neutron is \( E = \frac{1}{2} m v^2 \), so the velocity is: \[ v = \sqrt{\frac{2E}{m}}. \] Thus, the wavelength is: \[ \lambda = \frac{h}{\sqrt{2mE}}. \] Substituting the values of \( h = 6.626 \times 10^{-34} \, \text{J.s} \), and the mass of a neutron \( m \approx 1.675 \times 10^{-27} \, \text{kg} \): \[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 4.14 \times 10^{-21}}} \approx 1.56 \times 10^{-10} \, \text{m} = 0.156 \, \text{nm}. \] - Step 3: Energy of a photon having the same wavelength. The energy of a photon \( E_{\text{photon}} \) is related to the wavelength \( \lambda \) by: \[ E_{\text{photon}} = \frac{h c}{\lambda}, \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light. Substituting the values: \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.156 \times 10^{-9}} \approx 1.27 \times 10^{-16} \, \text{J}. \]