Special Relativity and Particle Decay

[TheTallOne]

Homework Statement

A particle with rest mass M_0 can decay at rest into a pair of particles each with rest
mass m_0. Calculate the following in the rest inertial frame of the original particle
using M_0c^2=600MeV and m_0c^2=150 MeV .
(i) The total energy of each particle produced in the decay.
(ii) The magnitude of the relativistic linear momentum of each particle produced in
the decay.
(iii) The speed of the particles produced in the decay.

Homework Equations

E^2=p^2c^2 + E_0^2 [1]
E=\gamma m_0c^2 [2]

The Attempt at a Solution

i)Ok, first I'm assuming that the new particles will have equal energy, so each E=300MeV
ii) I think I'm meant to use equation [1], where E=300 and E₀=150. This gives p=\frac{259.8}{c}=0.866. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?
iii) I use the fact that E=\gamma m_0c^2, giving \gamma=2 \Rightarrow v=\frac{\sqrt{3}}{2}c

Thanks very much for helping!

 

[ideasrule]

ii) I think I'm meant to use equation [1], where E=300 and E₀=150. This gives p=\frac{259.8}{c}=0.866. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?

That's correct. You can work out the units: E^2-E0^2 has units of MeV^2, so p has the unit MeV*s/m. To get the result in the familiar kg*m/s, just convert MeV to J before doing the calculation.

 

[vela]

Homework Statement

A particle with rest mass M_0 can decay at rest into a pair of particles each with rest
mass m_0. Calculate the following in the rest inertial frame of the original particle
using M_0c^2=600MeV and m_0c^2=150 MeV .
(i) The total energy of each particle produced in the decay.
(ii) The magnitude of the relativistic linear momentum of each particle produced in
the decay.
(iii) The speed of the particles produced in the decay.

Homework Equations

E^2=p^2c^2 + E_0^2 [1]
E=\gamma m_0c^2 [2]

The Attempt at a Solution

i)Ok, first I'm assuming that the new particles will have equal energy, so each E=300MeV.

Good. You're not really assuming the answer though; you're deducing it from symmetry.

ii) I think I'm meant to use equation [1], where E=300 and E₀=150. This gives p=\frac{259.8}{c}=0.866. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?

The method is correct. Just write the units in explicitly, and you'll see how they turn out.

(pc)^2 = E^2 - E_0^2 \rightarrow pc = \sqrt{(300~\textrm{MeV})^2 - (150~\textrm{MeV})^2} = 259.8~\textrm{MeV}

So your answer is in units of MeV/c.

iii) I use the fact that E=\gamma m_0c^2, giving \gamma=2 \Rightarrow v=\frac{\sqrt{3}}{2}c

Thanks very much for helping!

Since you have the momentum, you can also find the velocity using β=v/c=pc/E. Either method is fine, though.