Question

A neutron initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of 1.4 × 10⁻²⁶ kg-m/s and the antineutrino 6.4 × 10⁻²⁷ kg-m/s. Find the recoil speed of the proton (a) if the electron and the antineutrino are ejected along the same direction and (b) if they are ejected along perpendicular directions. Mass of the proton = 1.67 × 10⁻²⁷ kg.

Answer 1

It is given that:
Mass of proton, m_p = 1.67 × 10⁻²⁷ kg
Momentum of electron = 1.4 × 10⁻²⁶ kg m/s
Momentum of antineutrino = 6.4 × 10⁻²⁷ kg m/s

Let the recoil speed of the proton be V_p.

(a) When the electron and the antineutrino are ejected in the same direction,

Applying the law of conservation of momentum, we get:

$$\begin{gathered} m_p{V}_p=p_{\mathrm{electron}}+p_{\mathrm{antineutrino}} \\ 1.67\times {10}^{-27}\times V_{\mathrm{P}}=1.4\times {10}^{-26}+6.4\times {10}^{-27} \\ \Rightarrow V_{\mathrm{P}}=\left(\frac{20.4}{1.67}\right)=12.2\mathrm{m}/\mathrm{s},\mathrm{in}\mathrm{the}\mathrm{opposite}\mathrm{direction} \end{gathered}$$

(b) When the electron and the antineutrino are ejected perpendicular to each other,

Total momentum of electron and antineutrino is given as,

$$\begin{gathered} p=\sqrt{p_{\mathrm{electron}}+p_{\mathrm{antineutrino}}}=\sqrt{(14{)}^2+(6.4{)}^2}\times {10}^{-27}=15.4\times {10}^{-27}\mathrm{kg}\mathrm{m}/\mathrm{s} \\ \Rightarrow V_p=\frac{15.4\times {10}^{-27}\mathrm{kgm}/\mathrm{s}}{1.67\times {10}^{-27}\mathrm{kg}} \\ \Rightarrow V_p=9.2\mathrm{m}/\mathrm{s} \end{gathered}$$