Special Relativity: Photon emission by a moving atom

AI Thread Summary
The discussion centers on a physics problem involving an atom in an excited state emitting a photon while moving towards a scintillation counter. The key equations involve conservation of energy and momentum, with the initial energy in the lab frame being (Q_0 + mc^2)γ. The participant struggles with incorporating the Lorentz factor γ and simplifying the equations to derive the expected photon energy formula, Q = Q_0(1 + Q_0/(2mc^2)). A crucial point is that the excited state energy Q_0 is considered in the atom's rest frame, which affects the calculations. The discussion emphasizes the importance of correctly applying relativistic principles to solve the problem.
Helgi
Messages
2
Reaction score
0

Homework Statement


The problem involves an atom (Said to be in an excited state of energy Q_0) traveling towards a scintillation counter with speed v. The atom then emits a photon of energy Q and stops completely. The rest mass of the atom is m. I'm supposed to show that
Q = Q_0(1+\frac{Q_0}{2mc^2})

Homework Equations


It's mostly just conservation of energy and momentum stuff.
The kinetic energy of the atom is
K = (\gamma -1)m_0 c^2
I have a feeling this formula
E^2 = (cp)^2+m_0^2c^4
has to be used but I don't see where.

The Attempt at a Solution


What I figured I should do was to use conservation of energy and momentum. So I set up 2 equations

Q_0 + (\gamma-1)m_0c^2 = m_0 c^2+Q Which is the energy conservation
\gamma\cdot m_0v = \frac{Q}{C} and the conservation of momentum

Which I'm pretty sure is set up alright but when I try to solve for Q I don't get anything that is simplifies to what it's supposed to be. The \gamma and v usually get in the way and they're not supposed to be in the answer.

Now what I tried was to just simplify the first formula, and I ended up with
2Q_0 = 2m_c^2 + Q
this looks pretty reasonable but it's missing the second power on the Q_0.
 
Last edited:
Physics news on Phys.org
Helgi said:
The problem involves an atom (Said to be in an excited state of energy Q_0) traveling towards a scintillation counter with speed v. The atom then emits a photon of energy Q and stops completely. The rest mass of the atom is m. I'm supposed to show that
Q = Q_0(1+\frac{Q_0}{2mc^2})

The key point to note here that the excited state of energy Q0 is in the rest frame of the atom, not the lab frame as you have considered. This energy Q0 makes the rest energy of the atom (Q0+mc^2). So, the initial energy of the system in labframe is (Q0+mc^2)\gamma, and the initial momentum is (Q0/c^2+m)\gamma v.

The rest is as you have done. It's a two liner. Put c=1 to keep it neat, and put it back in the end to make it dimensionally correct.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
13
Views
2K
Replies
4
Views
1K
Replies
28
Views
4K
Replies
2
Views
902
Replies
38
Views
6K
Replies
1
Views
2K
Back
Top