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

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(A) The kinetic energy of a neutron at temperature $T$ is given by $K.E. = \frac{3}{2}kT = \frac{p^2}{2m}$.Thus,the momentum $p = \sqrt{3mkT}$.The de

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$\lambda /2$

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(A) The kinetic energy of a neutron at temperature $T$ is given by $K.E. = \frac{3}{2}kT = \frac{p^2}{2m}$.Thus,the momentum $p = \sqrt{3mkT}$.The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}$.This implies $\lambda \propto \frac{1}{\sqrt{T}}$.Given $T_1 = 27^{\circ}C = 300 \ K$ and $T_2 = 927^{\circ}C = 1200 \ K$.Therefore,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Hence,$\lambda_2 = \frac{\lambda}{2}$.

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

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