Ncert Exercises & Solutions

Before the neutrino hypothesis, the beta decay process was thought to be the transition,

$n \to p + \bar{e}$

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.

Hint: Use the law of conservation of momentum.

Step 1: Conserve the momentum.

Before

β

-decay, the neutron is at rest. Hence,

E_{n} = m_{n} c^{2}, p_{n} = 0

p_{n} = p_{p} + p_{e}

or p_{p} + p_{e} = 0 \Rightarrow |p_{p}| = |p_{e}| = p

Also, E_{p} = {(m_{p}^{2} c^{4} + {p_{p}}^{2} c^{2})}^{\frac{1}{2}}, E_{e} {= (m_{e}^{2} c^{4} + {p_{e}}^{2} c^{2})}^{\frac{1}{2}}

Step 2: Conserve the energy.

From the conservation of energy,

{(m_{p}^{2} c^{4} + p^{2} c^{2})}^{\frac{1}{2}} + {(m_{e}^{2} c^{4} + p^{2} c^{2})}^{\frac{1}{2}} = m_{n} c^{2}

m_{p} c^{2} \approx 936 MeV, m_{n} c^{2} \approx 938 MeV, m_{e} c^{2} = 0.51 MeV

Step 3: Find the energy of the electron and the proton.

Since the energy difference between n and p is small, pc will be small,

p c < < < m_{p} c^{2}

, while pc may be greater than

m_{e} c^{2}

\Rightarrow m_{p} c^{2} + \frac{p^{2} c^{2}}{2 m_{p}^{2} c^{4}} ≃ m_{n} c^{2} - p c

To first order p c ≃ m_{n} c^{2} - m_{p} c^{2} = 938 MeV - 936 MeV = 2 MeV

This gives the momentum of proton or neutron. Then,

E_{p} = {(m_{p}^{2} c^{4} + p^{2} c^{2})}^{\frac{1}{2}} = \sqrt{936^{2} + 2^{2}}

≃ 936 MeV

E_{e} = {(m_{e}^{2} c^{4} + p^{2} c^{2})}^{\frac{1}{2}} = \sqrt{{(0.51)}^{2} + 2^{2}}

≃ 2.06 MeV

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions