Question

The de-Broglie wavelength of a neutron at ${27}^{\circ }\text{C}$ is $\lambda$. The de-Broglie wavelength of the neutron at ${927}^{\circ }\text{C}$ will be

• A. $\frac{\lambda }{9}$
• B. $\frac{\lambda }{4}$
• C. $\frac{\lambda }{3}$
• D. $\frac{\lambda }{2}$

Answer 1

The correct option is D $\frac{\lambda }{2}$

The de-Broglie wavelength of the thermal neutron is,

$\lambda =\frac{h}{\sqrt{2mE}}$ , But $E=\frac{3kT}{2}$

$\lambda \propto \frac{1}{\sqrt{E}}\propto \frac{1}{\sqrt{T}}$

$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{273+27}{273+927}}$

$=\sqrt{\frac{300}{1200}}=\frac{1}{2}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.