Atom Emitting Photon: Mass Decrease Explained

In summary, according to the homework, when an atom emits a photon, its mass decreases. This is because energy and momentum are conserved.
  • #1
Welshy
20
0

Homework Statement


An atom initially at rest emits a photon. Explain why the mass of the atom decreases, taking into account energy and momentum conservation.


The Attempt at a Solution


I get the energy conservation part. E=delta(m)c^2=hf. But how would momentum conservation cause an atom at rest to decrease in mass? Unless it's talking about the zeroth component of a 4-vector - in which case that's just energy, surely?
 
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  • #2
Delta m c^2 = h f is not exactly valid, because then you are ignoring the change in momentum of the atom. If you pretend that the atom is in rest before and after the emission of the photon, then the change in energy of the atom is delta m c^2 and conservation of energy implies that this equals up to a minus sign h f.

You can say that the intial energy of the system is m c^2 and the final energy is gamma m' c^2 + hf. Now, we can write

gamma m' c^2 = m' c^2 + E_k

We know that E_k > 0 due to recoil. Energy conservation then implies that:

m c^2 = m' c^2 + E_k + h f

Then it follows that m' < m

Good exercise: Compute m' in terms of m and f
 
  • #3
Ok. My explanation of the mass deficit due to energy conservation was oversimplified. But the problem lies in the momentum conservation. Any idea how I would explain that?
 
  • #4
Welshy said:
Ok. My explanation of the mass deficit due to energy conservation was oversimplified. But the problem lies in the momentum conservation. Any idea how I would explain that?

You simply have to say that the photon has a momentum of hf/c, so the atom must have the same mometum, but in the opposite direction, This then means that the kinetic energy E_k is nonzero. Then since:

m c^2 = m' c^2 + E_k + h f

You have:


m' = m - [E_k + hf]/c^2

The contents of the square brackets is positive so this proves that
m' < m

But I think it is far more interesting to actually compute m'. That is a good exercise to practice the algebra of four vectors.
 
  • #5
Duh! I couldn't see the wood for the trees! :P Thanks!
 

1. What is an atom emitting photon?

An atom emitting photon is a process in which an atom releases energy in the form of a particle of light called a photon.

2. How does an atom emit a photon?

An atom emits a photon when one of its electrons transitions from a higher energy level to a lower energy level. This release of energy causes the electron to emit a photon.

3. What is the relationship between mass and photon emission?

The process of an atom emitting a photon does not result in a decrease in the mass of the atom. While the atom releases energy in the form of a photon, the total mass of the atom remains constant.

4. Why is it important to understand the process of atom emitting photon?

Understanding the process of atom emitting photon is important because it is a fundamental process in the behavior of matter and light. It helps us understand the behavior of atoms and their interactions with light, which has various applications in fields such as medicine, technology, and energy.

5. Can the mass decrease of an atom emitting photon be measured?

Yes, the mass decrease of an atom emitting photon can be measured using scientific instruments and techniques such as mass spectrometry. These methods can detect the subtle changes in mass that occur during photon emission.

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