In beta decay, a nucleus emits an electron. A bismuth-210 nucleus initially at r

Question

In beta decay, a nucleus emits an electron. A bismuth-210 nucleus initially at rest undergoes beta decay to polonium-210. [Perhaps new information for you - this equation is written as^210_83Bi rightarrow^210_84Po +^0_-1beta and this process occurs with a half-life of 5.12 days.] The emitted electron moves to the right with momentum of 5.60 times 10^-22 kg m/s. The Po-210 nucleus, with mass of 3.50 times 10^-25 kg recoils to the left with at a speed of 1.14 times 10^3 m/s. With the given information, demonstrate that momentum is not conserved. [The fact that momentum was not conserved allowed physicists to look for an unknown particle later named the antineutrino.] What is the magnitude and direction of the antineutrino momentum necessary for the system to conserve momentum?

Explanation / Answer

Given that

   beta decay of Bi to Po

as the electron moving towards right with momentum Pe = 5.60*10^-22 kgm/s

   and the recoil Po of mass m = 3.50*10^-22 kg, moving towards left with speed V_p = 1.14*10^3 m/s

now the momentum of Po is P_Po = m*V_p = 3.50*10^-25*1.14*10^3= 3.99*10^-22 kg.m/s

P_e = 5.60*10^-22 kgm/s
P_Po = 3.99*10^-22 kg m /s

so the momentum was not conserved

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