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

Question

In beta decay, a nucleus emits an electron. A ^210^Bi (bismuth) nucleus at rest undergoes beta decay to ^210^Po (polonium). Suppose that the emitted electron moves to the right with a momentum of (5.34 x 10^22) kg*m/s. The ^210^Po, with mass (2.78 x 10^25) kg, recoils to the left at a speed of (1.25 x 10^3) m/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 E 10).

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 E 28  kg*m/s. 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 E 28  kg*m/s.

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