Special Relativity, AP French 6.10: Atom Emits Photon

[raving_lunatic]

Homework Statement

I'll copy down the exact phrasing, so there's no question that this is what we're being asked:

"An atom in an excited state of energy Qo (as measured in its rest frame) above the ground state moves towards a scintillation counter with speed v. The atom decays to its ground state by emitting a photon of energy Q, as recorded by the counter, coming completely to rest as it does so. If the rest mass of the atom is M, show that Q = Qo[1 + Qo/2Mc² ]

Now, I wasn't sure if the relative motion between the emitter (atom) and observer (scintillation counter) might have some impact on the energy measured by the counter, or - if we are meant to take account of that - how exactly to do so.

Homework Equations

E² = (pc)² + (m₀c²)² -- can be applied to the atom before and after

Q = pc
for the photon

E = γm₀c² --- relativistic energy for a particle of rest mass m₀

The Attempt at a Solution

My first thought was to use conservation of energy and momentum in the rest frame of the atom.

This gives us energy before = m₀c² + Q₀ as the atom is at an energy Q₀ above the ground state as measured in its rest frame.

Energy after = γm₀c² + Q - that is the relativistic energy of the atom, which recoils at a velocity v in its rest frame (because it is at rest in the lab frame) added to the energy of the photon it emits.

We can also use conservation of momentum; this is a zero-momentum frame, so the total momentum has to be zero before and after the photon emission.

Therefore Q/c (photon momentum) + p (atom momentum) = 0

I tried to combine these to give the following expression:

m₀c² + Q₀ = Q + [(pc)² + (m₀c²)²]^0.5

But this didn't lead to the required expression, as I ended up with some unwanted cross terms QQ₀.

This is one of about five or six attempts along the same lines, and another unsuccessful attempt in the Lab Frame, and I'm beginning to think I must be missing something: every time, I can get an expression fairly close to what we're being required to show, but with some additional, unwanted cross terms.

I'd appreciate any guidance! Thank you :)

 

[tiny-tim]

hi raving_lunatic! welcome to pf!

My first thought was to use conservation of energy and momentum in the rest frame of the atom.
…
Energy after = γm₀c² + Q

no, Q (of the photon) is measured in the rest-frame of the atom, so you'll need to adjust it

 

[raving_lunatic]

I'm a little confused. Are you saying that Q is measured in the lab frame by the counter, and so when we're dealing with it in the atom's rest frame, we need to transform it using the Lorentz transformation?

I.e

Q' = γ(1-v/c)Q?

So we'd then have:

m₀c² + Q₀ = γm₀c² + γ(1-v/c)Q

 

[tiny-tim]

hi raving_lunatic!

I'm a little confused. Are you saying that Q is measured in the lab frame by the counter, and so when we're dealing with it in the atom's rest frame, we need to transform it using the Lorentz transformation?

I.e

Q' = γ(1-v/c)Q?

yes (what is worrying you about that? )

the question gives you measurements in two different frames, so you have to convert all to the same frame

(and similarly for the momentum equation!)

 

[raving_lunatic]

I think I have it now! Thank you for your help