A stationary radioactive nucleus of mass number 120 undergoes α-decay and releases a total energy of 30 MeV. Assume the daughter and α-particle are non-relativistic and their masses are proportional to their mass numbers. If the kinetic energy of the α-particle is N MeV, find N.
Physics · Nuclear physics · radioactivity and law of decay
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Question

A stationary radioactive nucleus of mass number 120 undergoes α-decay and releases a total energy of 30 MeV. Assume the daughter and α-particle are non-relativistic and their masses are proportional to their mass numbers. If the kinetic energy of the α-particle is N MeV, find N.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

After α-decay, the daughter mass number is 120−4 = 116. The parent was initially at rest, so momentum conservation gives equal and opposite momenta for daughter and α-particle: |p(α)| = |p(D)| = p. For the same momentum, K = p² ÷ (2M), so kinetic energy is inversely proportional to mass. Thus, K(α) : K(D) = M(D) : M(α) = 116 : 4 = 29 : 1. The total released energy is Q = K(α)+K(D) = 30 MeV. The 30 MeV is therefore divided into 30 equal parts, of which 29 parts go to the α-particle. K(α) = 29 Me…