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Nuclei — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsNuclei5 Marks
Question
Before the neutrino hypothesis, the beta decay process was throught to be the transition,
$\text{n}\rightarrow\ \text{p}+\overline{\text{e}}$
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.
✓

Answer

Let us consider the cases before and after $\beta-\text{decay}$
Before $\beta-\text{decay};$ if the neutron was at rest.
Hence $, E_n = m_nc^2, p_n = 0$
After $\beta-\text{decay};\text{p}_\text{n}=\text{p}_\text{p}+\text{p}_\text{e}$
or, $0=\text{p}_\text{p}+\text{p}_\text{e}$
$\Rightarrow\ |\text{p}_\text{p}|=|\text{p}_\text{e}|=\text{p}$
Also, $\text{E}_\text{p}=\big(\text{m}_\text{p}^2\text{c}^4+\text{p}_\text{p}^2\text{c}^2\big)^\frac{1}{2}$
$\text{E}_\text{e}=\big(\text{m}_\text{e}^2\text{c}^4+\text{p}_\text{p}^2\text{c}^2\big)^\frac{1}{2}=\big(\text{m}_\text{e}^2\text{c}^4+\text{p}_\text{e}^2\text{c}^2\big)^\frac{1}{2}$
Now applying conservation of energy,
$\big(\text{m}_\text{e}^2\text{c}^4+\text{p}^2\text{c}^2\big)^\frac{1}{2}=\big(\text{m}_\text{e}^2\text{c}^4+\text{p}^2\text{c}^2\big)^\frac{1}{2}=\text{m}_\text{n}\text{c}^2$
$\text{m}_\text{p}\text{c}^2\approx936\ \text{MeV},\text{m}_\text{e}\text{c}^2\approx938\ \text{MeV}$ and $\text{m}_\text{e}\text{c}^2=0.51\ \text{MeV}$
Since, the energy difference between $n$ and $p$ is small $, pc$ will be small $, pc < < < m_pc^2,$ while pc may be greater than $m_ec^2.$
$\Rightarrow\ \text{m}_\text{p}\text{c}^2+\frac{\text{p}^2\text{c}^2}{2\text{m}^2_\text{p}\text{c}^4}\simeq\text{m}_\text{n}\text{c}^2-\text{pc}$
To first otder $pc = m_nc^2 - m_pc^2 = 938\ \text{MeV} - 936\ \text{MeV} = 2\ \text{MeV}.$
This given the momentum of proton or neutron. Then,
$\text{E}_\text{p}=\big(\text{m}_\text{p}^2\text{c}^4+\text{p}^2\text{c}^2\big)^\frac{1}{2}=\sqrt{936^2+2^2}\approx936\ \text{MeV}$
$\text{E}_\text{e}=\big(\text{m}_\text{e}^2\text{c}^4+\text{p}^2\text{c}^2\big)^\frac{1}{2}=\sqrt{(0.51)^2+2^2}= 2.06\ \text{MeV}$

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