Question

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, $\frac{{\lambda }_1}{{\lambda }_2}$ is

• A. $\frac{m_1}{m_2}$
• B. $\frac{m_2}{m_1}$
• C. $1.0$
• D. $\frac{\sqrt{m_2}}{\sqrt{m_1}}$

Answer 1

The correct option is C $1.0$

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