Question

de-broglie wavelength of a neutron at 927 degree Celsius is lambda. what will be its wavelength at 27-degree Celsius?

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

Answer 1

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

We know

${\lambda }_{neutron}\propto \frac{1}{\sqrt{T}}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_2}{T_1}}$

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{(273+927)}{(273+27)}}$

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

$=2$

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