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  4. The de-Broglie wavelength of a neutron at 27° C is λ. What will be its wavelength at 927° C

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Q. The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda$. What will be its wavelength at $927^{\circ} C$

Solution:

$\lambda_{\text {neutron }} \propto \frac{1}{\sqrt{T}}$
$\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow \frac{\lambda}{\lambda_{2}}=\sqrt{\frac{(273+927)}{(273+27)}}$
$=\sqrt{\frac{1200}{300}}=2$
$\Rightarrow \lambda_{2}=\frac{\lambda}{2}$