# Nuclear Excitation

1. Feb 16, 2014

### stumpoman

1. The problem statement, all variables and given/known data
A nucleus of mass m initially at rest absorbs a gamma ray (photon) and is excited to a higher energy state such that its mass is now 1.01 m.

Find the energy of the incoming photon needed to carry out this excitation.

2. Relevant equations

$m^{2} = E^{2}-P^{2}$ possibly

$E_{sys}= KE+E_{o}+E_{\gamma}$

3. The attempt at a solution

Equating energy before and after.

$m+E_{\gamma}=1.01m+KE$

$E_{\gamma}=.01m+KE$

Is there a way to find the exact energy required? I think it may have something to do with the momentum but they do not give any info about the motion of the final particle.

2. Feb 16, 2014

### Dick

The motion of the final particle is fixed by conservation of momentum. They don't have to specify it.

3. Feb 16, 2014

### stumpoman

So the momentum before and after is E_gamma? I still have that kinetic energy value that I cannot do anything with.

4. Feb 16, 2014

### Dick

Sure it is. At least in units where c=1. Use $m^{2} = E^{2}-P^{2}$ on the final state.

5. Feb 16, 2014

### stumpoman

I think I may have solved it. I made a diagram.

Using the second portion

$(E_{\gamma}+m)^{2}-(E_{\gamma})^{2}=(1.01m)^{2}$

$2E_{\gamma}+m^{2}=1.0201m^{2}$

$E_{\gamma}=0.51005m$

This seems to make sense but I am not sure if it is correct.

#### Attached Files:

• ###### lorentz vectors.png
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6. Feb 16, 2014

### Dick

I'm not really familiar with that kind of diagram. But the solution looks correct.