# Homework Help: Particle Production

1. Mar 15, 2014

### JSGandora

1. The problem statement, all variables and given/known data
Particle A (mass 1000 MeV/c^2) is at rest at the center of a spherical gamma detector, which completely surrounds it except for a small hole. We accelerate particle B (mass 500 MeV/c^2) to a total energy of 700 MeV, sending it through the hole towards particle A. When the particles collide, a single, excited particle of unknown mass, C*, is produced. C* quickly decays to its unexcited state, C (mass 1300 MeV/c^2), emitting a single gamma ray.

a) Find the mass of C*.
b) If the gamma is emitted directly away from the hole, what energy does the detector record?
c) The experiment is repeated, and the detector records an energy of 300 MeV. At what angles did particle C and the gamma emerge? Assume that C* decays at the center of the detector. Let an angle of 0 be directly away from the hole, and an angle of $\pi$ be directly towards the hole.

2. Relevant equations
$E^2 = p^2 c^2 + m^2 c^4$

3. The attempt at a solution
a) We have the total energy of the system to be 1000 MeV/c^2+700 MeV/c^2 = 1700 MeV/c^2. We also have the total momentum of the system to be only from particle B which is

$p=\sqrt{ E_B^2/c^2 -m_B^2 c^2}=\sqrt{(700 MeV)^2/c^2 - (500 MeV)^2 / c^2} = 490 MeV/c$

Then by conservation of energy and momentum, the calculated momentum and energies do not change so the new particle C* has a mass of

$m^2=\sqrt{E^2/c^4-p^2 / c^2} =\sqrt{(1700MeV/c^2)^2 - (490 MeV/c^2)^2} = \boxed{1628 MeV/c^2}$

Can someone tell me if this is right or not?

b) It is given that the C* particle decays into its unexcited state C. Does that mean it decays into a state of 0 kinetic energy? So the gamma particle carries away an energy that is the difference of the total energy (1700 MeV) and the rest energy of C (given as 1300 MeV). So would the energy detector detect an energy of $1700 MeV - 1300 MeV = \boxed{400 MeV}$. But that would violate conservation of momentum which would require the sum of the momentum of the gamma ray $E_{\gamma}/c = 400 MeV/c$ and the momentum of the C particle $p_C=\sqrt{ E_C^2/c^2 -m_C^2 c^2} = 0$ (no velocity) to be equal to the momentum calculated before which was $490 MeV/c$ which is not true. However, if we solve the equations
$E_{\gamma}/c +\sqrt{ E_C^2/c^2 -m_C^2 c^2}=490 MeV/c$
and
$E_{\gamma} + E_C = 1700 MeV$
we get $E_C = 1303 MeV$ and $E_\gamma = 397 MeV$. So is the answer 397 MeV?

I also have no idea how to solve part d. Can someone help me? Thanks in advance.

2. Mar 15, 2014

### Staff: Mentor

The formulas are right, checking numbers is something for a computer.

In which frame? (=no)

The second approach looks good.

For part (c), your momenta need directions.

3. Mar 15, 2014

### JSGandora

If you change to the rest frame of C*, then only $E^2-p^2c^2=m^2 c^4$ is conserved which is, $1628 MeV$. Then if it decays to its unexcited state, we can apply conservation of energy (energy stays at 1628 MeV) and conservation of momentum (which stays at 0) to get

$E_\gamma + E_C = 1628 MeV$​
Momentum of Gamma ray: $\frac{E_\gamma}{c}$
Momentum of C particle: $\frac{\sqrt{E_C^2-m^2c^4}}{c}$ where $m=1300 MeV/c^2$.

And they must have velocities in the opposite direction in order to have a total momentum 0 so we have

$\frac{E_\gamma}{c}=\frac{\sqrt{E_C^2-(1300 MeV)^2}}{c}$​

The system has no real solutions according to wolfram alpha. I double checked my calculations from the first part and I don't think I have anything incorrect. Are my equations incorrect?

4. Mar 15, 2014

### Staff: Mentor

5. Mar 15, 2014

### JSGandora

Oh strange, I must have inputted it incorrectly. Then $E_C = 1333 MeV$. How do I convert that back to the initial reference frame?

6. Mar 16, 2014

### Staff: Mentor

You'll need the motion of the rest frame of C*, and some variable for the angle, and then a Lorentz transformation. I would use Eγ, however, as you know the photon energy in the lab frame.