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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 $\frac{m}{n}$ , then the minimum value of $m+n \, $ is

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
As there is no external force
Linear momentum will be conserved
$\textit{P}_{1}=\textit{P}_{2}=\textit{P}$
Now, de-Broglie wavelength
$\lambda _{1}=\frac{h}{P_{1}}\text{ or }\lambda _{2}=\frac{h}{P_{2}}$
$\Rightarrow $ $\frac{\lambda _{1}}{\lambda _{2}}=\frac{\textit{P}_{2}}{\textit{P}_{1}}=\frac{\textit{P}}{\textit{P}}=1$ $=\frac{m}{n}$
$\Rightarrow m+n=1+1=2$