Solving Particle Physics Relativity Problem: hbar*omega = DE (1 - DE/2Ea)

[Somethingsin]

Here's the problem:

" A particle is in its first excited state with energy Ea, before it decays to its ground state with energy Eg by emitting a photon with an energy hbar*omega

(omega = 2Pi*frequency)

Why will the photon energy be smaller than DE= Ea - E?

Show that:
hbar*omega = DE (1 - DE/2Ea)"

Now the first part seems easy enough. The photon has a momentum, so the particle must get a momentum as well, equal and opposite, so a part of the energy will be used to give the particle that momentum.

The second part is a little harder though... I have no idea.

So far I've managed:
hbar*omega = DE ( 1 - EKinetic/DE)
hbar*omega = DE (1 - (gamma-1)Eg/DE)
hbar*omega = DE ( 1- (DE + ((2gamma*Eg*Ea -2EgEa)/DE))/2Ea)
But I'm not so sure that is correct.


Any help would be most welcome

(Ps. I'm only first year, so please keep it reasonably comprehensible please)

 

[Andrew Mason]

Here's the problem:

" A particle is in its first excited state with energy Ea, before it decays to its ground state with energy Eg by emitting a photon with an energy hbar*omega

(omega = 2Pi*frequency)

Why will the photon energy be smaller than DE= Ea - E?

Show that:
hbar*omega = DE (1 - DE/2Ea)"

Now the first part seems easy enough. The photon has a momentum, so the particle must get a momentum as well, equal and opposite, so a part of the energy will be used to give the particle that momentum.

Correct. And the key to the second part is to quantify the energy used to give the particle that recoil momentum.

The second part is a little harder though... I have no idea.

So far I've managed:
hbar*omega = DE ( 1 - EKinetic/DE)
hbar*omega = DE (1 - (gamma-1)Eg/DE)
hbar*omega = DE ( 1- (DE + ((2gamma*Eg*Ea -2EgEa)/DE))/2Ea)
But I'm not so sure that is correct.

The momentum of the photon is: $p_{photon} = h/\lambda = \hbar \omega/c$

Since this is equal to the recoil momentum of the particle, the intial recoil speed of the particle is $v = p_{photon}/m = \hbar\omega/cm$.
Since $.5mv^2 + \hbar\omega = DE$, we have:

$$\frac{(\hbar\omega)^2}{2c^2m} + \hbar\omega - DE = 0$$

That is a quadratic equation in terms of $\hbar\omega$.
See if the solution to that gives you the result they are looking for.

AM