Physics - Modern Physics Question with Solution | TestHub

The de-Broglie wavelength of a particle in thermal equilibrium at temperature $T$ is given by $\lambda =\frac{h}{\sqrt{2mE}}$.

Since the kinetic energy $E=\frac{3}{2}kT$, we get $\lambda =\frac{h}{\sqrt{3mkT}}$.

Thus, $\lambda \propto \frac{1}{\sqrt{T}}$.

Given $T_1=27^{\circ }\mathrm{C}=300\mathrm{K}$ and $T_2=927^{\circ }\mathrm{C}=1200\mathrm{K}$.

$\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_1}{T_2}}$

$\frac{{\lambda }_2}{{\lambda }_0}=\sqrt{\frac{300}{1200}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

${\lambda }_2=\frac{{\lambda }_0}{2}$

Answer: $\frac{{\lambda }_0}{2}$