- #1

Sswift

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## Homework Statement

Estimate the neutron energy needed to produce fission of

^{208}Pb. Is it likely that such neutrons woould be released in the resulting fission?

## Homework Equations

E

_{excitation}=Q

_{excitation}+T

_{n}

Q

_{excitation}=[m(

^{208}Pb)+m

_{n}-m(

^{209}Pb)]c

^{2}

V=(e

^{2}/4πε

_{o})(Z

_{1}Z

_{2}/r)

r=1.2fm[(A

_{1})

^{1/3}+(A

_{2})

^{1/3}]

## The Attempt at a Solution

I'm a little confused just on how to approach this problem. I know that in order to fission the excitation energy must be greater than the activation energy. But doesn't the activation energy depend on what the products of the fission reaction are, of which there is more than one possible combination?

Anyways my best guess was to just come up with a fission reaction that could possibly happen and since Z=50 and N=82 are nuclear shell magic numbers I came up with the following

^{208}Pb + n →

^{132}Sn +

^{75}Ge + 2n

using

Q = [m(

^{208}Pb)+m

_{n}-m(

^{132}Sn) - m(

^{75}Ge)- 2m

_{n}]

and

E

_{activation}= V - Q

I find the activation energy and then find the kinetic energy the neutron needs for the excitation energy to be greater than the activation energy.

My main problem is I don't know if I'm alright assuming the fission products I did. And as I look around on the internet I get the sense that the activation energy is a set value for a particular isotope, which greatly confuses me. If anyone could just kinda push me in the right direction that would be greatly appreciated!