# Conservation Laws and Symmetries Particle Physics

1. Jul 22, 2013

### highwayman1739

2) A π0 of kinetic energy 350 MeV decays in flight into 2 γ rays of equal energies. Determine the angles of the γ rays from the incident π0 direction.

Not sure where I am going wrong but my answer is not correct.

Energy of π0 meson
E = 350 MeV

Rest mass Energy of π0 meson
E0 = 135 MeV

Initial momentum of π0 meson
P = √E^2 – Eo^2 / c
P = √(350 MeV)^2 – (135 MeV)^2 / c
P = 322.9163978 MeV/c

Law of conservation of Energy
E = Egamma + Egamma
Egamma = E/2
Egamma = 350MeV/2
Egamma = 175 MeV

Angle of gamma ray
θ = cos-1(Pc / 2E)
θ = cos-1(322.9163978 MeV/c) c / (2)(175 MeV)
θ = cos-1(0.922618279)
θ = 22.688 degrees

2. Jul 22, 2013

### collinsmark

According to the problem statement, the kinetic energy is 350 MeV. The total energy, E, is the kinetic energy plus the rest mass energy.

It wouldn't hurt to use parentheses or brackets in your notation.

p = √(E^2 – Eo^2)/ c

or

p = √(E2 – Eo2)/ c

or

$p = \frac{\sqrt{E^2 - E_0^2}}{c}$

Here you need to add the rest mass energy to the kinetic energy to find the total energy, E.

3. Jul 22, 2013

### highwayman1739

If I add kinetic energy and rest energy (350+135 = 485 MeV) and use this value for E my momentum seems to be to large (p = 465.83). Unless I am making another mistake in the part below.
θ = cos-1(Pc / 2E)
θ = cos-1(465.83 MeV/c) c / (2)(175 MeV)
θ = cos-1(1.33)(which cannot be )

4. Jul 22, 2013

### collinsmark

So far that looks right to me. Specifically, I got 465.83 MeV/c.

That's almost right, but you're missing an additional factor of 2 in there somewhere. I see you've put one '2' in there already, presumably because each photon's contributes half of the total momentum in the x direction. But also remember that each photon contributes half the total energy too. I only see one of the factor of '2's in there.

----

There is another way to do this, which is the long way. Use the formula E = hc/λ, and solve the the wavelength of each photon (remember that each photon only contributes half the total energy. Then realize |p| = h/λ, and that the x-component of each photon's momentum, px = pcosθ = hcosθ/λ, contributes to half the total momentum of the original pion. (Don't worry about the momentums in the y-direction, they are equal and opposite, so they cancel.)

You'll end up with a cosθ = ratio, where the "ratio" is guaranteed to be be less than 1 (assuming the original pion had non-zero rest mass, which of course is true).

[Edit: corrected a mistake.]

Last edited: Jul 22, 2013