Question

De-Broglie wavelength of neutron at ${27}^{\circ }C$ is $\lambda$ and its wavelength at ${927}^{\circ }C$ is $n\lambda$, then n is (The kinetic energy of the neutron is due to the thermal energy)
[Write upto two places after the decimal]

Answer 1

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$, where $E=\frac{3}{2}KT$

$\lambda =\frac{h}{\sqrt{3mKT}}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{1200}{300}}=2$

${\lambda }_2=\frac{{\lambda }_1}{2}={\lambda }_2=0.5\lambda$