Question

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

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

Answer 1

Answer: (A)

de-Broglie wavelength $\lambda \propto \frac{1}{\sqrt{T}}$
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_2}{T_1}}$
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{273+927}{273+27}}$
$=\sqrt{\frac{1200}{300}}=2$
or ${\lambda }_2=\frac{\lambda }{2}$