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

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(C) The de-Broglie wavelength $\lambda$ of a neutron is given by $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $E_k$ is the kinetic energy.For a neutron in

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$\frac{\lambda_0}{2}$

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(C) The de-Broglie wavelength $\lambda$ of a neutron is given by $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $E_k$ is the kinetic energy.For a neutron in thermal equilibrium at temperature $T$,the average kinetic energy is $E_k = \frac{3}{2} k_B T$.Thus,$\lambda = \frac{h}{\sqrt{2m(\frac{3}{2} k_B T)}} = \frac{h}{\sqrt{3mk_B T}}$.This implies $\lambda \propto \frac{1}{\sqrt{T}}$.Given $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$ and $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.Therefore,$\frac{\lambda_2}{\lambda_0} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.So,$\lambda_2 = \frac{\lambda_0}{2}$.

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda_0$. What will be its wavelength at $927^{\circ} C$?

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