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

 

[quicknote]

I would approach this problem by first understanding the basic principles of particle physics and relativity. In this situation, we have a particle in its first excited state with energy Ea and it decays to its ground state with energy Eg by emitting a photon with energy hbar*omega. The question is why the photon energy is smaller than the difference in energy between the two states (DE= Ea - E)?

To understand this, we need to consider the conservation of energy and momentum in this system. When the particle decays to its ground state, it emits a photon, which has momentum and energy. According to the law of conservation of momentum, the total momentum of the system (particle + photon) must remain constant. This means that the particle must also gain some momentum in the opposite direction to the photon.

Now, let's consider the energy of the system. The energy of the particle in its first excited state is Ea, and after the decay, it will be in its ground state with energy Eg. The energy of the photon is hbar*omega. According to the law of conservation of energy, the total energy of the system must also remain constant. This means that the energy of the particle must decrease by an amount equal to the energy of the photon emitted.

This is where the equation hbar*omega = DE (1 - DE/2Ea) comes in. This equation shows that the energy of the photon (hbar*omega) is equal to the difference in energy between the two states (DE) multiplied by a factor (1 - DE/2Ea). This factor takes into account the decrease in energy of the particle due to its gain in momentum.

In other words, the particle loses some of its energy in the form of momentum, resulting in a decrease in its energy by a factor of DE/2Ea. This is why the photon energy is smaller than the difference in energy between the two states.

I hope this explanation helps you understand the problem better. As a first year student, it's great that you are already tackling complex problems like this. Keep up the good work and don't hesitate to ask for help or clarification when needed. Science is a collaborative effort and we all learn from each other. Good luck!