Kinetic energies of particle decay

In summary, the lambda hyperon decays into a neutron and pion. The kinetic energies of the neutron and pion are 939.6 and 1115.7 MeV/c^2, respectively.
  • #1
SonOfOle
42
0

Homework Statement


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

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


Homework Equations


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


The Attempt at a Solution



Using the relevant equations, there are two equations and two unknowns ([tex]v_{n}[/tex] and [tex]v_{\pi}[/tex]). However, all my algebra attempts failed here. Any ideas? Is the above what I should be starting with?
 
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  • #2
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?
 
Last edited:
  • #3
Vanadium 50 said:
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.
[tex] E_{i}= E_{f} [/tex]

[tex] E_{i}= m_{\Lambda} c^{2} [/tex]

[tex] E_{f}= \gamma_{n} m_{n} c^{2} + \gamma_{\pi} m_{\pi} c^{2} [/tex]

for the first equation, conservation of energy.

And

[tex] \gamma_{n} m_{n} v_{n} = - \gamma_{\pi} m_{\pi} v_{\pi} [/tex]

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

[tex] \gamma_{i} = \left( \sqrt{1- (\frac{v_{i}}{c})^{2} } \right) ^{-1} [/tex]
 
  • #4
The gammas aren't helping. You might want to find an equation without them.
 
  • #5
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.
 
  • #6
Ah, right. Good ol' [tex] E^2 = p^2 c^2 + m_{i}^{2} c^4 [/tex] 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.
 

1. What is the definition of kinetic energy in particle decay?

Kinetic energy is the energy possessed by a particle due to its motion.

2. How is kinetic energy related to the decay of particles?

In the process of particle decay, the parent particle breaks apart into smaller daughter particles and releases energy in the form of kinetic energy.

3. What factors affect the kinetic energy of particles in decay?

The mass and velocity of the parent particle, as well as the mass and velocity of the daughter particles, all play a role in determining the amount of kinetic energy released in particle decay.

4. Can the kinetic energy of particles in decay be calculated?

Yes, the kinetic energy can be calculated using the formula KE = 1/2 * mv^2, where m is the mass of the particle and v is its velocity.

5. How is the conservation of energy applied to particle decay?

The conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. In particle decay, the total energy of the system (including the kinetic energy) must remain constant before and after the decay process.

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