Nuclear Excitation: Find Energy of Incoming Photon

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The discussion focuses on calculating the energy of a photon required to excite a nucleus from rest to a higher energy state, increasing its mass from m to 1.01m. The approach involves equating the total energy before and after the photon absorption, leading to the equation Eγ = 0.01m + KE. Momentum conservation is also considered, with the final state momentum linked to the energy of the incoming photon. A derived equation using the relationship m² = E² - P² results in Eγ = 0.51005m, which appears to be a plausible solution. The validity of this calculation is affirmed by participants, indicating confidence in the derived energy value.
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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.
 
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stumpoman said:

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

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stumpoman said:
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.
 
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