(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

^{224}Ra -->^{220}Rn + [itex]\alpha[/itex]

^{224}Ra -->^{212}Pb +^{12}C

^{224}Ra -->^{210}Pb +^{14}C

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

_{88}^{225}Ra = 18.818_{86}^{220}Rn = 10.604

_{82}^{212}Pb = -7.557 [itex]\alpha[/itex] = 2.425

_{82}^{210}Pb = -14.743_{6}^{14}C = 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 e^{2}/4[itex]\pi[/itex][itex]\epsilon[/itex]=1.44MeV.fm)

2. Relevant equations

Q = (M_{intial}-M_{resultants})

V_{C}= Z_{1}Z_{2}e^{2}/ 4[itex]\pi[/itex][itex]\epsilon[/itex]r

3. The attempt at a solution

For values of Q

^{224}Ra -->^{220}Rn + [itex]\alpha[/itex]

Q = 18.818-10.604-2.425 = 5.8MeV

^{224}Ra -->^{212}Pb +^{12}C

Q = 18.818+7.557-0 = 26.4MeV

^{224}Ra -->^{210}Pb +^{14}C

Q = 18.818+14.743-3.020 = 30.5MeV

And for the second part:

V_{C}= Z_{1}Z_{2}e^{2}/ 4[itex]\pi[/itex][itex]\epsilon[/itex]r

^{224}Ra -->^{220}Rn + [itex]\alpha[/itex]

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

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

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