A particle of mass $M$ at rest decays into masses $m_1$ and $m_2$ with non-zero velocities. The ratio of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ of the particles is

The particle of mass $M$ is initially at rest, so its initial momentum is zero. By the law of conservation of linear momentum, the final momenta of the two particles $m_1$ and $m_2$ must be equal in magnitude and opposite in direction.

Let $p_1$ be the magnitude of the momentum of particle $m_1$ and $p_2$ be the magnitude of the momentum of particle $m_2$. Then $p_1 = p_2 = p$.

The de Broglie wavelength $\lambda$ of a particle is given by the relation $\lambda = \dfrac{h}{p}$, where $h$ is Planck's constant.

For the two particles, we have $\lambda_1 = \dfrac{h}{p_1}$ and $\lambda_2 = \dfrac{h}{p_2}$.

Since $p_1 = p_2$, it follows that $\lambda_1 = \lambda_2$.

Therefore, the ratio $\dfrac{\lambda_1}{\lambda_2} = 1$.

Answer: 1:1