In beta decay, a nucleus emits an electron. A 210Bi (bismuth) nucleus at rest un

Question

In beta decay, a nucleus emits an electron. A 210Bi (bismuth) nucleus at rest undergoes beta decay to 210Po (polonium). Suppose that the emitted electron moves to the right with a momentum of 5.34*10^22 kg*m/s. The 210Po, with mass 2.78 *10^25 kg, recoils to the left at a speed of 1.25*10^3m/s. Momentum conservation requires that a second particle, called an antineutrino, must also be emitted. Calculate the magnitude of the momentum of the antineutrino that is emitted in this decay.

Give your answer in kg*m/s in scientific notation to three significant digits (i.e. 1.15*10^-67).

Explanation / Answer

According to conservation of momentum, i.e P (initial) = P (final)       0 =  5.34*10^22 kg*m/s. - 2.78 *10^25* 1.25*10^3 kg*m/s. + P( antineutrino )      ( since take right positive)      P( antineutrino ) = 3.47*10^28  kg*m/s.

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