A neutron initially at rest, decays into a proton, an electron an… - Vidyadip

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^-26kg-m/s and the antineutrino 6.4 × 10^-27kg-m/s. Find the recoil speed of the proton:

If the electron and the antineutrino are ejected along the same direction.
If they are ejected along perpendicular directions. Mass of the proton = 1.67 × 10^-27kg.

✓

Answer

Mass of proton = 1.67 × 10^-27

Let ‘V_p’ be the velocity of proton.

Given, momentum of electron = 1.4 × 10^-26kg m/sec.

Given, momentum of antineutrino = 6.4 × 10^-27kg m/sec

The electron & the antineutrino are ejected in the same direction. As the total momentum is conserved the proton should be ejected in the opposite direction.

1.67 × 10^-27 × V_p = 1.4 × 10^-26 + 6.4 × 10^-27 = 20.4 × 10^-27

$\Rightarrow\text{V}_{\text{p}}=\Big(\frac{20.4}{1.67}\Big)=12.2\text{m}/\text{sec}$ in the opposite direction.

The electron & antineutrino are ejected $\perp$ to each other.

Total momentum of electron and antineutrino,

$=\sqrt{(14)^2+(6.4)^2}\times10^{-27}\text{kg}\ \ \text{m}/\text{s}=15.4\times10^{-27}\text{kg}\ \ \text{m}/\text{sec}$

Since, 1.67 × 10^-27, V_p = 15.4 × 10^-27kg m/s

So, V_p = 9.2m/s

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