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,since the initial particle is at rest,the total momentum before and after decay must be ze

Vedclass · Accepted Answer

$1.0$

Vedclass · Answer

(C) According to the law of conservation of linear momentum,since the initial particle is at rest,the total momentum before and after decay must be 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 of the momenta,we have $|p_1| = |p_2| = p$,where $p = m_1 v_1 = m_2 v_2$.The de-Broglie wavelength is given by $\lambda = h/p$.Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths are $\lambda_1 = h/p$ and $\lambda_2 = h/p$.Therefore,the ratio $\lambda_1/\lambda_2 = (h/p) / (h/p) = 1.0$.

$A$ 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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