Question

If the de Broglie wavelength of a neutron at a temperature of 77°C is \( \lambda \), then the de Broglie wavelength of the neutron at a temperature of 1127°C is:

• A. \( \frac{\lambda}{2} \)
• B. \( \frac{\lambda}{3} \)
• C. \( \frac{\lambda}{4} \)
• D. \( \frac{\lambda}{9} \)

Hint:

The de Broglie wavelength varies inversely with the square root of temperature: \( \lambda \propto \frac{1}{\sqrt{T}} \).
- Convert all temperatures to Kelvin before calculations.

Answer 1

The Correct Option is A

The de Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m k_B T}} \] where: - \( h \) is Planck’s constant, - \( p \) is the momentum, - \( m \) is the mass, - \( k_B \) is Boltzmann’s constant, - \( T \) is the absolute temperature in Kelvin. 1. Convert temperatures to Kelvin: \[ T_1 = 77 + 273 = 350 \text{ K} \] \[ T_2 = 1127 + 273 = 1400 \text{ K} \] 2. Ratio of wavelengths: Since \( \lambda \propto \frac{1}{\sqrt{T}} \), \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} \] \[ = \sqrt{\frac{350}{1400}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \] \[ \lambda_2 = \frac{\lambda_1}{2} \] Thus, the correct answer is \(\boxed{\frac{\lambda}{2}}\).