A particle of mass M at rest decays into two particles of masses $m_1$ and $m_2$, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, ${\lambda }_1/{\lambda }_2$ is

By law of conservation of momentum,

$0=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}\Rightarrow m_1\overrightarrow{v_1}=-m_2\overrightarrow{v_2}$

– ve sign indicates that both he particles are moving in opposite direction. Now de-Broglie wavelengths

${\lambda }_1=\frac{h}{m_1{v}_1}$ and ${\lambda }_2=\frac{h}{m_2{v}_2}$; $\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{m_2{v}_2}{m_1{v}_1}=1$