Hi, quick problem regarding emission of gamma ray.

In summary, a Fe (57) atom in an excited state 14.4 keV above the ground state decays to the ground state with the emission of a gamma ray. Despite having no mass, photons do have momentum and can be taken into account when using conservation of momentum to calculate the recoil speed of the nucleus. By setting the initial velocity of the excited nucleus to 0, we can approximate the recoil velocity using the equation 0 = mv + E/c, making the problem simpler than initially thought.
  • #1
teclo
117
0
A Fe (57) atom is in an excited state 14.4 keV above the ground state. The nucleus decays to the ground state with the emission of a gamm ray. What's the recoil speed of the nucleus?

I'm not sure how to set this up. I thought a photon would have no mass, therefor no momentum. If so I couldn't do this with conservation of momenumtum or energy. I must be missing something, anyone have any ideas?
 
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  • #2
The closest thing my book has to a simliar example is dealing with the Rutherford experiement.

K(1) + 0 = K(3) + K(4)

so i would think that the initial energy of the atom equals the energy of the photon plus the energy of the atom after emission. I'm not sure about how to setup something like this though.
 
  • #3
teclo said:
I thought a photon would have no mass, therefor no momentum.
Despite being massless, photons do have momentum: p = E/c.
 
  • #4
teclo said:
A Fe (57) atom is in an excited state 14.4 keV above the ground state. The nucleus decays to the ground state with the emission of a gamm ray. What's the recoil speed of the nucleus?

I'm not sure how to set this up. I thought a photon would have no mass, therefor no momentum. If so I couldn't do this with conservation of momenumtum or energy. I must be missing something, anyone have any ideas?

A photon DOES have a momentum.

http://scienceworld.wolfram.com/physics/Photon.html

Even the classical version of light as EM radiation has something called "radiation pressure".

So, knowing this, can you now do the problem?

Zz.
 
  • #5
ZapperZ said:
A photon DOES have a momentum.

http://scienceworld.wolfram.com/physics/Photon.html

Even the classical version of light as EM radiation has something called "radiation pressure".

So, knowing this, can you now do the problem?

Zz.

yes, acutally i worked the problem out earlier. i forgot that p of the photon is E/c.

with that it was rather easy to approximate the recoil velocity of the nucleus using conservation of momentum. if reference frame observes initial excited nucleus as v of 0

0 = mv + E/c

i was making this harder than it really was
 

1. What causes the emission of gamma rays?

Gamma rays are emitted by atoms that have excess energy in their nuclei. This excess energy can be caused by a variety of processes, such as radioactive decay, nuclear reactions, or extremely high energy collisions.

2. Are gamma rays harmful to humans?

Yes, high levels of exposure to gamma rays can be harmful to humans. They have the ability to penetrate deep into our body tissue and can cause damage to cells and DNA. However, small amounts of exposure from natural sources, such as the sun or certain foods, are not harmful.

3. How are gamma rays detected?

Gamma rays are detected using specialized equipment called gamma ray detectors. These detectors measure the energy and intensity of gamma rays and can be used to identify their source.

4. What are some practical uses of gamma rays?

Gamma rays have many practical uses in various industries. They are used in medical imaging to diagnose and treat diseases, in industrial processes to sterilize equipment and food, and in security systems to detect nuclear materials.

5. Can gamma rays be shielded or blocked?

Yes, gamma rays can be shielded or blocked by materials with a high atomic number, such as lead or concrete. The thickness of the shielding needed depends on the energy and intensity of the gamma rays.

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