Finding the recoil energy of an excited atom when it de-excites

In summary, the conversation discusses finding the recoil energy of an excited atom when it de-excites. Through momentum and energy conservation equations, it is determined that the recoil energy is equal to the energy of the emitted photon divided by two times the rest mass of the atom times the speed of light squared. This equation can be used to find the recoil energy of other particles as well.
  • #1
sadhu
157
0
I know that i am commiting some serious errors in this , but just can't notice them
suppose we have to find the recoil energy of an excited atom when it de excites

suppose the total energy of the process is E

an atom will emit a photon on returning to normal state . energy of the photon is... e

recoil velocity of the atom is v;
mass of the atom is M

Mv+e/c=0....momentum conservation...c=velocity of light.initialy the atomis at rest

Mv*v/2 + e=E...energy conservation


from first equation

v=-e/(Mc)...(i)

thus recoil energy must come out by substituting v

Me*e/(2*M*M*c*c)

e*e/(2Mc*c)

substituting this in energy conservation equation

e+e*e/(2*Mc*c)=E
e+e*e/(2*Mc*c)-E=0

solving for e;
[tex]\Delta[/tex]=1+2E/(M*c*c)

whats is the solution ??
 
Last edited:
Physics news on Phys.org
  • #2


The recoil energy (K) is given as:

[tex]K = E^{2}_{photon}/2mc^{2}[/tex]

where mc2 describes the rest mass of the particle. mc2 will be of the order 109 to 1011 eV for atoms and nuclei.
 
  • #3


The solution is correct, but there are a few errors in the notation and units.

Firstly, the unit for energy is usually measured in joules (J) or electron volts (eV), so it would be more appropriate to use these units instead of just "e". Additionally, in the momentum conservation equation, the units do not match up. The left side is in units of mass times velocity (kg*m/s) while the right side is in units of energy divided by the speed of light (J*s/m). It would be more accurate to write it as Mv = -e/c.

In the second equation, the units also do not match up. The left side is in units of mass times velocity squared (kg*m^2/s^2) while the right side is in units of energy (J/eV). It would be more accurate to write it as Mv^2/2 + e = E.

Additionally, it would be helpful to define the variables used in the equations. For example, it is not clear what "E" represents in this context. Is it the total energy of the atom or the energy of the photon emitted? Clarifying this would make the solution easier to understand.

Overall, the solution is correct but it would benefit from clearer notation and units. It is important to pay attention to these details in order to avoid errors and confusion.
 

1. How do you find the recoil energy of an excited atom when it de-excites?

The recoil energy of an excited atom when it de-excites can be found using the formula: E = (hν)^2 / 2mc^2, where h is the Planck's constant, ν is the frequency of the emitted photon, m is the mass of the atom, and c is the speed of light.

2. Can the recoil energy of an excited atom be measured in a laboratory setting?

Yes, the recoil energy of an excited atom can be measured in a laboratory setting by using specialized equipment such as a mass spectrometer or by analyzing the Doppler shift in the frequency of the emitted photon.

3. Does the recoil energy of an excited atom vary for different types of atoms?

Yes, the recoil energy of an excited atom can vary depending on the mass and characteristics of the atom. Heavier atoms will have a higher recoil energy compared to lighter atoms.

4. How does the recoil energy of an excited atom affect its motion?

The recoil energy of an excited atom can cause it to move in the opposite direction of the emitted photon with a velocity proportional to the recoil energy. This can result in a slight change in the atom's position and momentum.

5. Is there a limit to the recoil energy of an excited atom?

According to the laws of physics, there is no limit to the recoil energy of an excited atom. However, in practical situations, the recoil energy is usually very small and does not have a significant effect on the atom's motion.

Similar threads

Replies
6
Views
1K
  • Other Physics Topics
Replies
5
Views
835
  • Quantum Physics
2
Replies
38
Views
3K
Replies
1
Views
999
  • Other Physics Topics
Replies
8
Views
1K
  • Other Physics Topics
Replies
1
Views
979
Replies
8
Views
760
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Special and General Relativity
4
Replies
131
Views
9K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top