Question

Before the neutrino hypothesis, the beta decay process was throught to be the transition, `n -> p + vece`. If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.

Answer 1

Before β decay, neutron is at rest. Hence E_n = m_nc², p_n = 0

After β decay, from conservation of momentum:

p_n = p_p + p_e

Or p_p + p_e = 0

⇒ |p_p| = |p_e| = p

Also, `E_p = (m_p^2c^4 + p_p^2c^2)^(1/2)`,

`E_e = (m_e^2c^4 + p_e^2c^2)^(1/2) = (m_e^2c^4 + p_pc^2)^(1/2)`

From the conservation of energy: 

`(m_p^2c^4 + p^2c^2)^(1/2) + (m_e^2c^4 + p^2c^2)^(1/2) = m_nc^2`

`m_pc^2 ≈ 936 MeV, m_nc^2 ≈ 938 MeV, m_ec^2 = 0.51 MeV`

Since the energy difference between n and p is small, pc will be small, PC << m_pc², while pc may be greater than m_ec².

⇒ `m_pc^2 + (p^2c^2)/(2m_p^2c^4) ≃ m_nc^2 - pc`

To first order `pc ≃ m_nc^2 - m_pc^2` = 938 MeV – 936 Mev = 2 MeV

This gives momentum. Then, 

`E_p = (m_p^2c^4 + p^2c^2)^(1/2) = sqrt(936^2 + 2^2) ≃ 936  MeV`

`E_e = (m_e^2c^4 + p^2c^2)^(1/2) = sqrt((0.51)^2 + 2^2) ≃ 2.06  MeV`