Neutron energy needed to produce fission of 208Pb

AI Thread Summary
To estimate the neutron energy needed for the fission of 208Pb, it is essential to determine the activation energy, which varies based on the fission products. The proposed fission reaction is 208Pb + n → 132Sn + 75Ge + 2n, and the activation energy can be calculated using the equation Eactivation = V - Q, where Q is derived from the masses of the involved isotopes. The discussion highlights the confusion around whether the activation energy is a fixed value for a specific isotope and emphasizes the relationship between the activation energy and the binding energy of the last neutron in 209Bi. If the activation energy is less than the binding energy, fission can occur with a neutron at rest; otherwise, additional kinetic energy is required. Understanding these relationships is crucial for accurately estimating the neutron energy needed for fission.
Sswift
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Homework Statement



Estimate the neutron energy needed to produce fission of 208Pb. Is it likely that such neutrons woould be released in the resulting fission?

Homework Equations



Eexcitation=Qexcitation+Tn
Qexcitation=[m(208Pb)+mn-m(209Pb)]c2
V=(e2/4πεo)(Z1Z2/r)
r=1.2fm[(A1)1/3+(A2)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

208Pb + n → 132Sn + 75Ge + 2n

using
Q = [m(208Pb)+mn-m(132Sn) - m(75Ge)- 2mn]

and
Eactivation= 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!
 
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Sswift said:

Homework Statement



Estimate the neutron energy needed to produce fission of 208Pb. Is it likely that such neutrons woould be released in the resulting fission?

Homework Equations



Eexcitation=Qexcitation+Tn
Qexcitation=[m(208Pb)+mn-m(209Pb)]c2
V=(e2/4πεo)(Z1Z2/r)
r=1.2fm[(A1)1/3+(A2)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

208Pb + n → 132Sn + 75Ge + 2n

using
Q = [m(208Pb)+mn-m(132Sn) - m(75Ge)- 2mn]

and
Eactivation= 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!

It's correct that Eactivation= V - Q
if Eactivation less than binding energy of last neutron in 209Bi
then a neutron with zero kinetic energy , fission immediately occur
but if Eactivation greater than binding energy of last neutron
the difference between them is neutron kinetic energy
 
also see John .R lamarsh- introduction to nuclear reactor theory chapter 3
 
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