Question

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

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

Answer 1

The correct option is D 1:1

By momentum conservation, $Mv=m_1{v}_1+m_2{v}_2...\left(1\right)$

As M is initially rest so $v=0$

Now (1) becomes, $0=m_1{v}_1+m_2{v}_2$

or $m_1{v}_1=-m_2{v}_2$, it means that the magnitude of momentum of both masses are equal i.e $p_1=p_2$

From de Broglie hypothesis, ${\lambda }_1=h/p_1$ and ${\lambda }_2=h/p_2$

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