Kinetic Energies of Neutron and Pion After Lambda Hyperon Decay?

[SonOfOle]

Homework Statement

A lambda hyperon \Lambda^{0} (mass = 1115.7 MeV/c^2) at rest in the lab frame decays into a neutron n (mass = 939.6 MeV/c^2) and a pion \pi^{0}: \Lambda^{0} \rightarrow n + \pi^{0}

What are the kinetic energies (in the lab frame) of the neutron and pion after the decay?

Homework Equations

KE = m c^{2} - m_{0} c^{2} = m_{0} c^{2} (\gamma - 1)
Conservation of Momentum: p_{n} = - p_{\pi}
p = \gamma m v

The Attempt at a Solution

Using the relevant equations, there are two equations and two unknowns (v_{n} and v_{\pi}). However, all my algebra attempts failed here. Any ideas? Is the above what I should be starting with?

 

[Vanadium 50]

Your first "relevant equation" isn't particularly relevant.

Note that the pion and proton masses aren't anywhere in your equations. Does this seem right to you?

 

[SonOfOle]

Your first "relevant equation" isn't particularly relevant.

Note that the pion and proton masses aren't anywhere in your equations. Does this seem right to you?

I suppose I should've filled in more what I implied by the first equation.
E_{i}= E_{f}

E_{i}= m_{\Lambda} c^{2}

E_{f}= \gamma_{n} m_{n} c^{2} + \gamma_{\pi} m_{\pi} c^{2}

for the first equation, conservation of energy.

And

\gamma_{n} m_{n} v_{n} = - \gamma_{\pi} m_{\pi} v_{\pi}

for the conservation of momentum. Those are the two equations and two unknowns I used, where

\gamma_{i} = \left( \sqrt{1- (\frac{v_{i}}{c})^{2} } \right) ^{-1}

 

[Vanadium 50]

The gammas aren't helping. You might want to find an equation without them.

 

[Vanadium 50]

Or rather...the gammas aren't helping right away.

You need to start by getting one equation in one unknown. What would be the most useful thing to solve for? Hint: it's common to both decay particles.

 

[SonOfOle]

Ah, right. Good ol' E^2 = p^2 c^2 + m_{i}^{2} c^4 The momentums are equal so can set both p's to be the same, we know the initial energy, we know the masses, solve for momentum, etc...

Not pretty math, but it works out.

Thanks Vanadium 50.