Question

A neutron initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of 1.4x10^(-26) kg-m/s and the antineutrino 6.4x10^(-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 = 1.67x10^(-27) kg.

Answer 1

Thanks for asking the question!

ANSWER::

Mass of proton = 1.67x10⁻²⁷ kg

Let 'V₀' be the velocity of proton.

Given:-
Momentum of electron = 1.4x10⁻²⁶ kg m/s
Momentum of antineutrino = 6.4 x 10⁻²⁷ kg m/s

(a)Electron and antineutrino are ejected in same direction and as the total momentum is conserved the proton should be ejected in opposite direction.

1.67 x 10⁻²⁷ x V₀ = 1.4 x 10⁻²⁸ + 6.4 x 10⁻²⁷

1.67 x 10⁻²⁷ x V₀ = 20.4 x 10⁻²⁷

V₀ = (20.4 x 10⁻²⁷) / (1.67 x 10⁻²⁷) = 12.2 m/s

So , velocity of proton is 12.2 m/s in opposite direction.

(b) Electron and antineutrino are ejected perpendicular to each other.

Total momentum of electron and antineutrino = √[(14)² + (6.4)²] x 10⁻²⁷
                                                                        = 15.4 x 10⁻²⁷ kg m/s

So , 1.67 x 10⁻²⁷ x V₀ = 15.4 x 10⁻²⁷ kg m/s

V₀ = 9.2 m/s

Hope it helps!