Question

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

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

Answer 1

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

de-Broglie wavelength is given by,

$\lambda =\frac{h}{p}$$\Rightarrow \lambda \propto \frac{1}{p}$

$\because \text{Average K.E. of thermal neutrons}=\frac{3}{2}kT$

$KE=\frac{3}{2}kT$ $\Rightarrow \frac{1}{2}\frac{p^2}{m}=\frac{3}{2}kT$

$\Rightarrow p=\sqrt{3mkT}$

$\Rightarrow \lambda \propto \frac{1}{p}\propto \frac{1}{\sqrt{T}}$

Given, $T_1=27+273=300\text{K};T_2=927+273=1200\text{K}$

$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{300}{1200}}=\frac{1}{2}$

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

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