If λ be the de Broglie wavelength of a thermal neutron at 27°C, then the wavelength of the same neut

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If λ be the de Broglie wavelength of a thermal neutron at 27°C, then the wavelength of the same neutron at 927°C is

Accepted Answer

$\lambda \propto \frac{1}{v} \propto \frac{1}{\sqrt{T}} \Rightarrow \frac{\lambda_{2}}{\lambda_{1}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{\lambda_{2}}{\lambda_{1}} = \sqrt{\frac{300}{1200}}$

$\therefore\lambda_{2} = \frac{\lambda}{2}$

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