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 ${\lambda }_1/{\lambda }_2$ is :

• A. $m_1/m_2$
• B. $m_2/m_1$
• C. $1$
• D. $\sqrt{m_2}/\sqrt{m_1}$

Answer 1

Answer: (C)

$1$

By conservation of momentum, $Mv=m_1{v}_1+m_2{v}_2$

As mass M is at rest so $v=0$. Thus, $m_1{v}_1=-m_2{v}_2$, it indicates the magnitude of momentum of masses is same i.e $p_1=p_2$

So, de Broglie wavelengths for masses $m_1$ and $m_2$ are ${\lambda }_1=h/p_1$ and ${\lambda }_2=h/p_2$

Thus, $\frac{{\lambda }_1}{{\lambda }_2}=\frac{p_2}{p_1}=1$ as $(p_1=p_2)$