I'm writing a simulation to model alpha-decay, and I need to know the kinetic energies of the alpha particles that are emitted. In tables, e.g. for the decay of 212-Polonium, you would read

$^{212}$Po -10.3649 MeV

where the value given is the mass excess $\Delta M$. According to a nuclear physics book (Krane), the kinetic energy of the alpha particle is

$$K = \frac{Q}{1 + m_\alpha / m_{x'}}$$

where $Q$ is the Q-value, which is "the amount of energy released in the reaction", $m_\alpha$ is the mass of the alpha particle and $m_{x'}$ is the mass of the daughter nucleus.

I thought that $Q$ and $\Delta M$ were simply related by a factor of $c^2$, $$Q = (\Delta M) c^2.$$ However, this gives the wrong answer. How can I calculate the kinetic energy of the alpha particle?


1 Answer 1


Hint :You have masses (from parent nuclei mass you can get mass of daughter nuclei by subtracting mass of $\alpha$ particle , and the Q value ie. the energy that gets liberated

and $931.5 \ MeV \approx 1 amu$ and so you can see the excess mass is easily negligible ie $<0.1amu$

$^{212}Po \rightarrow \ ^{208}D + \ ^4\alpha$

momentum is zero before and after the disintegration. => $m_D V_D=m_\alpha V_\alpha$

So , net energy $$1/2m_DV_d^2+1/2m_\alpha V_\alpha^2=8.95412MeV$$ $$1/2m_dV_d^2=1/2V_d(m_\alpha V\alpha)=1/2\dfrac{m_\alpha V_\alpha}{m_d}\times (m_\alpha V\alpha)=\dfrac{m_\alpha}{m_d}\times KE_\alpha$$ so, $$\Bigg(1+\dfrac{m_\alpha}{m_d}\Bigg)\times KE_\alpha=8.95412MeV$$ Now solve this to get the KE of $\alpha$ particle

  • $\begingroup$ thanks for the answer! according to this, then the last equation would just read $m_\alpha*V_\alpha^2 = 2 Ekin = 10.3649$ MeV, i.e. $E_{kin}^\alpha \approx 5$ Mev? which means that we are assuming that the Nucleus does not recoil? $\endgroup$
    – seb
    Commented Apr 5, 2013 at 11:05
  • $\begingroup$ No.$1/2m_dV_d^2=1/2V_d(m_\alpha V\alpha)=1/2\dfrac{m_\alpha V_\alpha}{m_d}\times (m_\alpha V\alpha)=\dfrac{m_\alpha}{m_d}\times KE_\alpha$ , and u need to solve it. $\endgroup$
    – ABC
    Commented Apr 5, 2013 at 11:09
  • $\begingroup$ sorry, I have been sitting behind the screen for too long ;-) $\endgroup$
    – seb
    Commented Apr 5, 2013 at 11:10
  • $\begingroup$ >@sebastian And now you can see it's same as in your book. $\endgroup$
    – ABC
    Commented Apr 5, 2013 at 11:14
  • 2
    $\begingroup$ in German we say "I had plie-wood in front of my eyes". of course! thank you! $\endgroup$
    – seb
    Commented Apr 5, 2013 at 11:15

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