Radioactive decay: nuclear physics, kinetics and relativity

Problem

A nucleus of $^{210}$Po (mass $\approx 209{,}982\;u$) decays by $\alpha$ emission into $^{206}$Pb (mass $\approx 205{,}974\;u$) plus an $\alpha$ particle (mass $\approx 4{,}0026\;u$). The initial nucleus is at rest. Determine: (a) the energy released, (b) the velocity of the $\alpha$ particle, (c) whether relativistic treatment is necessary.

Solution

(a) Energy released (mass defect). The energy comes from the mass that “disappears” in the decay (Einstein’s mass-energy equivalence). The mass defect is: $\Delta m = m(^{210}\text{Po}) - m(^{206}\text{Pb}) - m(\alpha) = 209{,}982 - 205{,}974 - 4{,}0026 = 0{,}0054\;u$ Converting with $1\;u \approx 931{,}5$ MeV$/c^2$: $\ev{Q = \Delta m\, c^2 \approx 0{,}0054 \times 931{,}5 \approx 5{,}0\;\text{MeV}}$

(b) Velocity of the $\alpha$ particle. The initial nucleus is at rest, so the total momentum is conserved and equal to zero: the two fragments recoil with opposite momenta, $p_\alpha = p_\text{Pb}$. The energy $Q$ is shared out inversely proportional to the masses, so the (lighter) $\alpha$ particle takes the larger share: $E_\alpha \approx Q\,\frac{m_\text{Pb}}{m_\text{Pb}+m_\alpha} = 5{,}0\times\frac{205{,}974}{209{,}977} \approx 4{,}93\;\text{MeV}$ I use the classical kinetic formula $E_\alpha = \tfrac12 m_\alpha v_\alpha^2$ (I will check afterwards that this is valid). With $m_\alpha = 4{,}0026\,u = 6{,}65\cdot 10^{-27}\;$kg and $E_\alpha = 4{,}93\;\text{MeV} = 7{,}90\cdot 10^{-13}\;$J: $v_\alpha = \sqrt{\frac{2E_\alpha}{m_\alpha}} = \sqrt{\frac{2\cdot 7{,}90\cdot 10^{-13}}{6{,}65\cdot 10^{-27}}} \approx 1{,}5\cdot 10^{7}\;\text{m/s}$ $\ev{v_\alpha \approx 1{,}5\cdot 10^{7}\;\text{m/s}}$

(c) Is relativity needed? The ratio with the speed of light is $\frac{v_\alpha}{c} \approx \frac{1{,}5\cdot 10^{7}}{3{,}0\cdot 10^{8}} \approx 0{,}05$ that is, about 5% of $c$. The Lorentz factor differs from unity by barely $\sim 0{,}1\%$: the classical formula $E_k=\tfrac12 m v^2$ is more than adequate. Relativity, on the other hand, is essential in point (a), where $Q = \Delta m\,c^2$: without mass-energy equivalence the decay could not be explained at all.

The connections: nuclear physics (ch. 22) for the mass defect, conservation of momentum (ch. 5) for the energy split, mass-energy equivalence (ch. 20) for the very source of $Q$.

Links

Topics: Nuclear physics · Special relativity Concepts: Radioactive decay · Nuclear binding energy · Mass-energy equivalence · Conservation of momentum