Question

A radioactive nucleus decays by $\beta$ emission. Both $\beta$ and neutrino move mutually at right angle with momentum $6\times {10}^{-21}kgm{s}^{-1}$ and $3\times {10}^{-21}kgm{s}^{-1}$. The direction of recoil of nucleus with respect to electron is :

• A. $ta{n}^{-1}\left(\frac{1}{2}\right)$
• B. $ta{n}^{-1}\left(2\right)$
• C. $180-ta{n}^{-1}\left(\frac{1}{2}\right)$
• D. $180-ta{n}^{-1}\left(2\right)$

Answer 1

The correct option is C $180-ta{n}^{-1}\left(\frac{1}{2}\right)$

As the nucleus was at rest, its final momentum must equal the opposite of the vector sum of the electron & neutrino momenta,.which are at right angles

Let the electron direction be along the positive x-axis and neutrino be -y axis

magnitude of nucleus momentum is $p=\sqrt{6^2+3^2}\times {10}^{-21}=\sqrt{45}\times {10}^{-21}kgm{s}^{-1}$

direction is $-ta{n}^{-1}\left(\frac{3\times {10}^{-21}}{6\times {10}^{-21}}\right)=-ta{n}^{-1}\left(\frac{1}{2}\right)$

Therefore the recoil direction is opposite, which is $180-ta{n}^{-1}\left(\frac{1}{2}\right)$