Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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

share|improve this answer
    
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? –  lomppi 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. –  Mr.ØØ7 Apr 5 '13 at 11:09
    
sorry, I have been sitting behind the screen for too long ;-) –  lomppi Apr 5 '13 at 11:10
    
thank, you got it! –  lomppi Apr 5 '13 at 11:14
1  
in German we say "I had plie-wood in front of my eyes". of course! thank you! –  lomppi Apr 5 '13 at 11:15
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.