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Admittedly, Nuclear Physics is not my strength. I'm writing a simulation to model alpha-decay. So far, I have looked up the values of the kinetic energy of the alpha particles that are emitted in a certain decay. Now I have seen this a lot of times, for instance for 212-Polonium:

$^{212}$Po -10.3649 MeV

the the value given is the mass excess. Then I looked into a Nuclear Physics book (Krane) and found:

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

where $T_\alpha$ is the kinetic energy of the alpha particle, Q is the Q-value, $m_\alpha$ is the mass of the alpha particle and $m_{x'}$ is the mass of the daughter nucleus.

First I thought, Q and $\Delta M$ are simply related by a factor of $c^2$. But that does not seem to be the case, since my calculations are wrong.

How can I calculate the kinetic energy of the alpha-particle only given the information above?

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up vote 7 down vote accepted

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

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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? – seb Apr 5 '13 at 11:05
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. – ABC Apr 5 '13 at 11:09
sorry, I have been sitting behind the screen for too long ;-) – seb Apr 5 '13 at 11:10
thank, you got it! – seb Apr 5 '13 at 11:14
in German we say "I had plie-wood in front of my eyes". of course! thank you! – seb Apr 5 '13 at 11:15

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