In beta decay, a nucleus emits an electron A ²¹⁰Bi (bismuth) nucleus at rest undergoes beta decay to ²¹⁰Po (polonium). Suppose the electron moves to the right with a momentum of 5.60 X 10^-22kg.m/s. The ²¹⁰Po nucleus, with mass 3.50 X 10^-25 kg, recoils to the left at a speed of 1.14 X 10³ m/s. Momentum conservation requires that a second particle, called an antineutrino, must also be emitted. Calculate the magnitude and direction of the antineutrino that is emitted in this decay.

The given data is listed below as-

• The mass of the Polonium nucleus is $m_c=3.5\times {10}^{-25}kg.$,
• The velocity of the polonium nucleus is$V_c=-1.14\times {10}^{3}m/s$
• The momentum of the electron is ${\rho }_A=5.60\times {10}^{-22}\mathrm{kg}\cdot m{s}^{-1}$

The momentum of a particle is the product of the particle’s mass and velocity and is given by-

$\overrightarrow{\mathbf{\rho }}\mathbf{=}\mathbf{m}\overrightarrow{\mathbf{v}}$

Where m is the mass of the particle and V is the velocity of the particle.

Let the subscripts A, B and C indicate to the electron, antineutrino, and polonium nucleus, respectively. Let +x be to the right, so after the decay,

The velocity of the polonium nucleus will be $V_c=-1.14\times {10}^{3}m/s$

The momentum of the electron is

The total momentum of a system is conserved since there is no external net force on a system.

Therefore $\overrightarrow{\rho }=m\overrightarrow{v}$,

The Bismuth nucleus was at rest before the decay.

So, after the decay,

${\overrightarrow{\rho }}_A+{\overrightarrow{\rho }}_B+{\overrightarrow{\rho }}_B=0$

${\overrightarrow{\rho }}_B=-({\overrightarrow{\rho }}_A+{\overrightarrow{\rho }}_B)$…………(2)

Substitute m_c and V_c in equation (2)

Therefore,

$\begin{array}{c}{\overrightarrow{\rho }}_C=(3.50\times {10}^{-25})\cdot (1.14\times {10}^3)\\ =-3.99\times {10}^{-22}\mathrm{kg}\cdot m{s}^{-1}\end{array}$

Now substitute values of${\rho }_A$ and ${\rho }_C$in equation (2) to obtain

role="math" localid="1659338258799" $\begin{array}{c}{\overrightarrow{\rho }}_B=-(5.60\times {10}^{-22}-3.99\times {10}^{-22})\\ =-1.61\times {10}^{-22}\mathrm{kg}\cdot m{s}^{-1}\end{array}$

${\overrightarrow{\rho }}_B=1.61\times {10}^{-22}\mathrm{kg}\cdot m{s}^{-1}$to the left.

Thus, magnitude and direction of momentum of the anti neutrino is$1.61\times {10}^{-22}\mathrm{kg}\cdot m{s}^{-1}$ to the left.