# Homework Help: Finding Q values of decays and finding distance where Coulomb Barrier = Q value

1. Jan 21, 2012

### bmarson123

1. The problem statement, all variables and given/known data
224Ra --> 220Rn + $\alpha$
224Ra --> 212Pb + 12C
224Ra --> 210Pb + 14C

Calculate the Q-Values (in MeV) for these decays given their atomic mass excesses (in MeV) are

88225Ra = 18.818 86220Rn = 10.604
82212Pb = -7.557 $\alpha$ = 2.425
82210Pb = -14.743 614C = 3.020

For these 3 decays estimate the distance from the centre of the nucleus at which the Coulomb Barrier is equal to the calculated Q value (assuming e2/4$\pi$$\epsilon$=1.44MeV.fm)

2. Relevant equations

Q = (Mintial-Mresultants)

VC = Z1Z2e2/ 4$\pi$$\epsilon$r

3. The attempt at a solution

For values of Q

224Ra --> 220Rn + $\alpha$

Q = 18.818-10.604-2.425 = 5.8MeV

224Ra --> 212Pb + 12C

Q = 18.818+7.557-0 = 26.4MeV

224Ra --> 210Pb + 14C

Q = 18.818+14.743-3.020 = 30.5MeV

And for the second part:

VC = Z1Z2e2/ 4$\pi$$\epsilon$r

224Ra --> 220Rn + $\alpha$

5.8 = (88 x 82 x 1.44) /r
r = 1791.5 fm

I thought that this seemed quite a large value. Also, for the second 2 reactions I wasn't sure if this was the correct equation to be using because I thought it was specificaly just for alpha decay, and I didn't think the last 2 reactions were alpha.

Any pointers would be fantastic!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 22, 2012

### gomboc

It looks fine, and yes, that equation for Q-value should work for all nuclear decays.

There is one problem though: you seemed to have added the product masses instead of subtracting them in the calculation of the Q-values for the second and third decays.