Question
When the nucleus of a radium-226, which is at rest, decays, an $$\alpha $$ particle and the nucleus of radon are created. The released energy during the decay is $$4.87\,MeV,$$ which appears as the kinetic energy of the two resulted particles. Calculate the kinetic energy of $$\alpha $$ particle and radon nucleus.
A.
$$4.78\,MeV$$ and $$0.09\,MeV$$
B.
$$4.67\,MeV$$ and $$0.2\,MeV$$
C.
$$4.84\,MeV$$ and $$0.03\,MeV$$
D.
$$4.81\,MeV$$ and $$0.06\,MeV$$
Answer :
$$4.78\,MeV$$ and $$0.09\,MeV$$
Solution :
$${k_\alpha } = \frac{M}{{M + 4}}Q = \frac{{222}}{{226}} \times 4.87 = 4.78\,MeV$$