Question

The de-Broglie wavelength of a neutron at \( 27^\circ\text{C} \) is ' \( \lambda_0 \) '. What will be its wavelength at \( 927^\circ\text{C} \)?

• A. \( \frac{\lambda_0}{4} \)
• B. \( \frac{\lambda_0}{3} \)
• C. \( \frac{\lambda_0}{2} \)
• D. \( \frac{3\lambda_0}{2} \)

Hint:

Always convert temperature to Kelvin ($K$) before applying proportionality in Physics.

Answer 1

The Correct Option is C

Step 1: Formula
The de-Broglie wavelength of a thermal neutron is $\lambda = \frac{h}{\sqrt{3mkT}} \implies \lambda \propto \frac{1}{\sqrt{T}}$.
Step 2: Temperature in Kelvin
$T_1 = 27 + 273 = 300\text{ K}$.
$T_2 = 927 + 273 = 1200\text{ K}$.
Step 3: Ratio Calculation
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
$\lambda_2 = \frac{\lambda_0}{2}$.
Final Answer: (C)