Question

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:

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

Answer 1

Answer: (D)

1:1

Initial momentum of the particle is zero as it is at rest.

Thus according to law of conservation of momentum, the final momentum of the system must be zero.

Let the momentum of the two fragments be $p_1$ and $p_2$.

$⟹$ $p_1=p_2$

de-Broglie wavelength $\lambda =\frac{h}{p}$

where $p$ is the momentum of the particle and $h$ is Planck's constant.

$\therefore$ ${\lambda }_1=\frac{h}{p_1}$ and ${\lambda }_2=\frac{h}{p_2}$

$p_1=p_2$ $⟹$ ${\lambda }_1={\lambda }_2$