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

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(A) The de-Broglie wavelength of a neutron is given by $\lambda = \frac{h}{\sqrt{2mkT}}$,where $m$ is the mass of the neutron,$k$ is the Boltzmann con

Vedclass · Accepted Answer

$\lambda / 2$

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(A) The de-Broglie wavelength of a neutron is given by $\lambda = \frac{h}{\sqrt{2mkT}}$,where $m$ is the mass of the neutron,$k$ is the Boltzmann constant,and $T$ is the absolute temperature.Since $h$,$m$,and $k$ are constants,we have $\lambda \propto \frac{1}{\sqrt{T}}$.Therefore,$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$.Given $T_1 = 27^oC = 27 + 273 = 300 \ K$ and $T_2 = 927^oC = 927 + 273 = 1200 \ K$.Substituting the values: $\frac{\lambda}{\lambda_2} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.Thus,$\lambda_2 = \frac{\lambda}{2}$.

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

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