Q. The de-Broglie wavelength of a neutron at $927℃$ is $\lambda$. What will be its wavelength at $27℃$ ?
A) $\lambda/2$
B) $\lambda/4$
C) $4\lambda$
D) $2\lambda$
Explanation
(d) We know that, $\lambda = \frac{h}{\sqrt{2mkT}}$ ;
So, $\lambda \propto \frac{1}{\sqrt{T}}$
$\therefore\ \ \frac{\lambda_{27}}{\lambda_{927}} = \sqrt{\frac{927 + 273}{27 + 273}} = 2$
or $\lambda_{27} = 2\lambda_{927} = 2\lambda$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.