1. Oct 11, 2005

### 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)

2. Oct 11, 2005

### Andrew Mason

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

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