Question

Neutron decay in the free space is given as follow:
${}_0{n^1} \to {}_1{H^1} + {}_{ - 1}{e^0} + \left[ {} \right]$
The, the parenthesis represents?

Answer 1

Hint:Given problem related to radioactive decay. To solve such type of problems following basic points are necessary
1) Radioactivity (The phenomenon of spontaneous disintegrations)
2) ${\beta ^ + }$Decay and ${\beta ^ - }$ decay
3) Theory of neutrino \[(\nu \,\,\& \,\,\overline \nu )\]

Complete step by step answer:
For solving the given problem, first we know the basic theory of $\beta $ decay specially ${\beta ^ - }$ decay and ${\beta ^ + }$ decay.
${\beta ^ - }$ Decay: In ${\beta ^ - }$ decay a neutron decay into a proton and electron. Thus the proton number of the parent nucleus increases by one and the nucleon number remains unchanged. The electron escapes from the atom and leaves behind a positively charged atom.
${}_Z^AP \to {}_{Z + 1}^AP + {}_{ - 1}^0e + \left[ {\overline \nu } \right]$
Here $\overline \nu $ (antineutrino) and $\nu $ (neutrino)
The neutrino has zero electrical charge and negligible mass (not zero).
Remember: In $\beta $ decay
1) An electron and an antineutrino are emitted
2) A positron and a neutrino are emitted
From the given problem:
${}_0{n^1} \to {}_1{H^1} + {}_{ - 1}{e^0} + \left[ {} \right]$
In this equation ${}_0{n^1}$ decay into ${}_1{H^1}$and an electron and on the basis of conservation of momentum $\left[ {\overline \nu } \right]$ is emitted when ${\beta ^ - }$ decay occurs.

Note:Revise the complete theory of radioactivity. Carefully observe the decay reactions of $\alpha ,\beta ,\gamma $ decay. You should have the knowledge of neutrino and antineutrino