91Ó°ÊÓ

The given data is listed below as-

• The mass of the Polonium nucleus is $m_c=3.5×{10}^{−25}\text{ }kg.$,
• The velocity of the polonium nucleus is$V_c=−1.14×{10}^3\text{ }m/s$
• The momentum of the electron is ${\mathrm{ÒÏ}}_A=5.60×{10}^{−22}\text{ }\mathrm{kg}â‹\dots m{s}^{-1}$

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

$\stackrel{\mathbf{→}}{\mathbf{ÒÏ}}\mathbf{=}\mathbf{m}\stackrel{\mathbf{→}}{\mathbf{v}}$

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 $V_c=−1.14×{10}^3\text{ }m/s$

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 $\stackrel{→}{\mathrm{ÒÏ}}=m\stackrel{→}{v}$,

The Bismuth nucleus was at rest before the decay.

So, after the decay,

${\stackrel{→}{\mathrm{ÒÏ}}}_A+{\stackrel{→}{\mathrm{ÒÏ}}}_B+{\stackrel{→}{\mathrm{ÒÏ}}}_B=0$

${\stackrel{→}{\mathrm{ÒÏ}}}_B=−({\stackrel{→}{\mathrm{ÒÏ}}}_A+{\stackrel{→}{\mathrm{ÒÏ}}}_B)$…â¶Ä¦â¶Ä¦â¶Ä¦(2)

Substitute m_c and V_c in equation (2)

Therefore,

$\begin{array}{c}{\stackrel{→}{\mathrm{ÒÏ}}}_C=(3.50×{10}^{−25})â‹\dots (1.14×{10}^3)\\ =−3.99×{10}^{−22}\text{ }\mathrm{kg}â‹\dots m{s}^{-1}\end{array}$

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

role="math" localid="1659338258799" $\begin{array}{c}{\stackrel{→}{\mathrm{ÒÏ}}}_B=−(5.60×{10}^{−22}−3.99×{10}^{−22})\\ =−1.61×{10}^{−22}\text{ }\mathrm{kg}â‹\dots m{s}^{-1}\end{array}$

${\stackrel{→}{\mathrm{ÒÏ}}}_B=1.61×{10}^{−22}\text{ }\mathrm{kg}â‹\dots m{s}^{-1}$to the left.

Thus, magnitude and direction of momentum of the anti neutrino is$1.61×{10}^{−22}\text{ }\mathrm{kg}â‹\dots m{s}^{-1}$ to the left.