Question

Find the Kinetic energy of a neutron emerging as a result of the decay of a stationary ∑– hyperon (∑– → n + π–). ​

Answer 1

Topic :-

Atomic and Nuclear Physics

Given :-

A neutron gets emerged as a result of the decay of a stationary Σ⁻ hyperon.

$\sum^- \to n + \pi^-$

To Find :-

The Kinetic Energy of neutron emerged.

Concept Used :-

Total Energy of a relativistic particle,

$E=m_0c^2+T$

where,

T is Kinetic Energy of the particle.

Invariant used while examining collisions,

$E^2-p^2c^2=m_{0}^{2}c^4$

where,

E is total energy of the system prior to collision

p is total momentum of the system prior to collision

Solution :-

Σ⁻ is decaying into two particles.

From Momentum conservation,

$\vec {p}_n+\vec {p}_{\pi^-}=0$  

which represents that total momentum of the system prior to collision is Zero. ( p = 0 )

As we know,

$E^2-p^2c^2=m_{0}^{2}c^4$

$E^2-(0)^2c^2=m_{0}^{2}c^4$

$E^2=m_{0}^{2}c^4$

So, we can write,

$E_{\pi^-}^2-E_{n}^2=m_{\pi^-}^2c^4-m_{n}^2c^4$

From Energy conservation,

$m_{\sum^-}c^2=E_n+E_{\pi^-}$

$(m_{\sum^-}c^2-E_n)^2=(E_{\pi^-})^2$

$m^2_{\sum^-}c^4-2m_{\sum^-}c^2E_n+E^2_n=E_{\pi^-}^2$

$E_{\pi^-}^2-E^2_n=m^2_{\sum^-}c^4-2m_{\sum^-}c^2E_n$

So,

$m^2_{\sum^-}c^4-2m_{\sum^-}c^2E_n=m_{\pi^-}^2c^4-m_{n}^2c^4$

$E_n=\dfrac{m_{\sum^-}^2c^2+m_n^2c^2-m_{\pi^-}^2c^2}{2m_{\sum^-}}$

Now, apply formula for total energy,

$E_n=m_nc^2+T_n$

$T_n=E_n-m_nc^2$

$T_n=\left ( \dfrac{m_{\sum^-}^2+m_n^2-m_{\pi^-}^2}{2m_{\sum^-}} -m_n \right )c^2$

$T_n=\left ( \dfrac{(m_{\sum^-}-m_n)^2-m_{\pi^-}^2}{2m_{\sum^-}} \right )c^2$

Now, substitute the values,

$T_n=19.5\:\:MeV$

Answer :-

So, the Kinetic Energy of emerged neutron is 19.5 MeV.

Answer 2

Explanation:

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