Nuclear Excitation: Find Energy of Incoming Photon

[stumpoman]

Homework Statement

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.

Homework Equations

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

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

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.

 

[Dick]

Homework Statement

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.

Homework Equations

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

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

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.

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

 

[stumpoman]

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

 

[Dick]

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

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

 

[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.

 

[Dick]

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.

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