The ratio of de Broglie wavelengths associate | vedclass

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(D) The de Broglie wavelength $\lambda$ of a thermal neutron at temperature $T$ is given by the formula $\lambda = \frac{h}{\sqrt{3mkT}}$,where $h$ is

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$5: 4$

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(D) The de Broglie wavelength $\lambda$ of a thermal neutron at temperature $T$ is given by the formula $\lambda = \frac{h}{\sqrt{3mkT}}$,where $h$ is Planck's constant,$m$ is the mass of the neutron,and $k$ is the Boltzmann constant.From this relation,we see that $\lambda \propto \frac{1}{\sqrt{T}}$.Given temperatures are $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$ and $T_2 = 352^{\circ} C = 352 + 273 = 625 \ K$.The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values,$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{625}{400}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.Thus,the ratio is $5: 4$.

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

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