Question:medium

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

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