Question

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

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

Answer 1

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

$\frac{1}{2}m{V}^2=\frac{3}{2}kT$
$\Rightarrow mV=\sqrt{3mKT}$
$\lambda =\frac{h}{p}=\frac{h}{mV}=\frac{h}{\sqrt{3mKT}}$
$\Rightarrow \lambda \alpha =\frac{1}{\sqrt{T}}$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{273+927}{273+27}}=2$
$\Rightarrow {\lambda }_2=\frac{{\lambda }_1}{2}=\frac{\lambda }{2}.$