Question:
De-Broglie wavelength of neutron at \(27^\circ C\) is \(\lambda\). Find at \(927^\circ C\).
De-Broglie wavelength of neutron at \(27^\circ C\) is \(\lambda\). Find at \(927^\circ C\).
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De-Broglie wavelength $\propto$ \(1/\sqrt{T}\) for thermal particles.
Updated On: Apr 23, 2026
- \(\lambda/2\)
- \(\lambda/3\)
- \(\lambda/4\)
- \(\lambda/9\)
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The Correct Option is A
Solution and Explanation
Concept:
\[
\lambda \propto \frac{1}{\sqrt{T}}
\]
Step 1: Convert temperature
\[ T_1 = 27^\circ C = 300K,\quad T_2 = 927^\circ C = 1200K \]
Step 2: Ratio
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2} \] Conclusion: \[ \lambda_2 = \frac{\lambda}{2} \]
Step 1: Convert temperature
\[ T_1 = 27^\circ C = 300K,\quad T_2 = 927^\circ C = 1200K \]
Step 2: Ratio
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2} \] Conclusion: \[ \lambda_2 = \frac{\lambda}{2} \]
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