Q. A particle of mass M at rest decays into two masses $m_{1}$ and $m_{2}$ with non-zero velocities. The ratio of de-Broglie wavelengths of the particles $\frac{\lambda_{1}}{\lambda_{2}}$ is
A) $\frac{m_{2}}{m_{1}}$
B) $\frac{m_{1}}{m_{2}}$
C) $\frac{\sqrt{m_{1}}}{\sqrt{m_{2}}}$
D) 1:1
Explanation
(d) By law of conservation of linear momentum
$m_{1}v_{1} = m_{2}v_{2}$
So, $m_{1}v_{1} = m_{2}v_{2}$
Now, de-Broglie wavelength$\ \lambda = \frac{h}{mv}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} = \frac{m_{2}v_{2}}{m_{1}v_{1}}\ \ \ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda_{1}:\lambda_{2} = 1:1\ \ \ \ \ \ \ \ \ \ \ \ \ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.