Finding Q values of decays and finding distance where Coulomb Barrier = Q value

[bmarson123]

Homework Statement

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)

Homework Equations

Q = (M_intial-M_resultants)

V_C = Z₁Z₂e²/ 4$\pi$$\epsilon$r

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:

V_C = 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!

 

[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.