# Neutron energy needed to produce fission of 208Pb

• Sswift
In summary, to estimate the neutron energy needed for fission of 208Pb, the equation Eexcitation=Qexcitation+Tn can be used. The activation energy must be greater than the excitation energy for fission to occur. The activation energy depends on the fission products, which can vary. A possible fission reaction for 208Pb is 208Pb + n → 132Sn + 75Ge + 2n, using Q=[m(208Pb)+mn-m(132Sn)-m(75Ge)-2mn]. The activation energy can be found by subtracting Q from the potential energy V. The difference between the activation energy and the binding energy of the last neutron
Sswift

## 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!

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

## What is the definition of neutron energy?

Neutron energy refers to the kinetic energy of a neutron, which is the energy it possesses due to its motion.

## How is fission of 208Pb produced?

Fission of 208Pb is produced when a neutron collides with a nucleus of 208Pb, causing it to split into two smaller nuclei and releasing energy.

## What is the role of neutron energy in fission of 208Pb?

The neutron energy plays a crucial role in initiating the fission process by providing enough energy to overcome the strong nuclear force holding the nucleus together.

## What is the minimum neutron energy needed to produce fission of 208Pb?

The minimum neutron energy needed to produce fission of 208Pb is known as the fission threshold energy, which for 208Pb is approximately 1.9 MeV (million electron volts).

## How does the neutron energy affect the fission process of 208Pb?

The neutron energy directly affects the fission process by determining the probability of a successful fission event. Higher neutron energies have a greater chance of causing fission compared to lower energies.

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