Question

A radioactive nucleus initially at rest decays by emitting an electron and neutrino at right angles to one another. The momentum of the electron is $3.2\times {10}^{-23}kg-m/sec$ and that of the neutrino is $6.4\times {10}^{-23}kg-m/sec$. The direction of the recoiling nucleus with that of the electron motion is:

• A. ${\tan }^{-1}\left(0.5\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. $\pi -{\tan }^{-1}\left(2\right)$
• D. $\frac{\pi }{2}+{\tan }^{-1}\left(2\right)$

Answer 1

The correct option is C $\pi -{\tan }^{-1}\left(2\right)$

$\tan \theta =\frac{p_n}{p_e}=\frac{6.4\times {10}^{-23}}{3.2\times {10}^{-23}}=2$
According to law of conservation of momentum, the residual nucleus most move in a direction just opposite to that $p_R$. Hence direction of the recoiling nucleus with that of the electron motion
$\varphi =\pi -\theta =\pi -{\tan }^{-1}\left(2\right)$