Question

The de Broglie wavelength of a thermal neutron at ${927}^{o}C$ is $\lambda$. Its wavelength at ${327}^{o}C$ will be

• A. $\lambda /2$
• B. $\lambda /\sqrt{2}$
• C. $\lambda \sqrt{2}$
• D. $2\lambda$

Answer 1

Answer: (C)

$\lambda \sqrt{2}$

The relation between de Broglie wavelength and temperature is given by, $\lambda =\frac{h}{\sqrt{2mkT}}$

where $k=$ Boltzmann's constant , $m=$ mass, $T=$ temperature.

Here all are constant except $T$ so $\lambda \propto \frac{1}{\sqrt{T}}$

Thus, $\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{273+927}{273+327}}=\sqrt{2}$

or ${\lambda }_2=\lambda \sqrt{2}$