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)

$\frac{\lambda }{2}$

Answer 2

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 $\lambda_{2} =\frac{\lambda}{2}$