The de-Broglie wavelength of a neutron at 27^ | vedclass

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(A) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.Since $h$,$m$,and $

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$\lambda / 2$

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(A) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.Since $h$,$m$,and $k$ are constants,we have $\lambda \propto \frac{1}{\sqrt{T}}$.Given $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$ and $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}$,we get $\frac{\lambda_2}{\lambda} = \sqrt{\frac{300}{1200}}$.$\frac{\lambda_2}{\lambda} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$\lambda_2 = \frac{\lambda}{2}$.

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

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