A particle of mass M at rest decays into two | vedclass

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(C) According to the law of conservation of linear momentum,the initial momentum of the system is zero.$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2$$\Rightarrow

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(C) According to the law of conservation of linear momentum,the initial momentum of the system is zero.$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2$$\Rightarrow m_1 \vec{v}_1 = -m_2 \vec{v}_2$Taking the magnitude,we get $m_1 v_1 = m_2 v_2$,which implies that the magnitudes of the momenta of the two particles are equal $(p_1 = p_2 = p)$.The de-Broglie wavelength is given by $\lambda = h / p$.Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths will be $\lambda_1 = h / p$ and $\lambda_2 = h / p$.Therefore,the ratio $\lambda_1 / \lambda_2 = (h / p) / (h / p) = 1$.

$A$ 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

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