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

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