Question

The decay product of tritium is:
A. \[_1^1{\text{H}}\]
B.\[_1^2{\text{H}}\]
C.\[_2^3{\text{He}}\]
D.\[_2^4{\text{He}}\]

Answer 1

Hint:Negative \[\beta \] decay causes increase in atomic number by 1 and atomic mass remains the same. As the atomic number of Tritium is 1, it will increase by 1 and result in 2 which is atomic number of Helium.

Complete step by step answer:
Spontaneous emission of radiations from the nucleus is known as radioactivity. The rays which are emitted are namely alpha rays, beta rays and gamma rays. Their properties are as follow:

Features	\[\alpha \]	\[\beta \]	\[\gamma \]
Identity	Helium nucleus or double ionized helium atom \[_2{\text{H}}{{\text{e}}^4}\]	Fast moving electrons \[_ - {\beta ^0}{\text{ or }}{\beta ^ - }\]	EM wave
Charge	Twice of proton \[ + 2{\text{e}}\]	Electronic \[ - {\text{e}}\]	Neutral
Mass	4 times of mass of proton	Rest mass of electron	Rest mass is zero
Effect of electric or magnetic field	deflection	Deflection more than alpha	No deflection

Equation of \[\alpha \] decay: after \[\alpha \] emission, obtained daughter nucleus has 2p and 2n less than its parent nucleus. This is as follow: \[{}_{\text{Z}}{{\text{X}}^{\text{A}}} \to {}_{{\text{Z}} - 2}{{\text{Y}}^{{\text{A}} - 4}} + {}_2{\text{H}}{{\text{e}}^4}\] .\[\beta \] Decay is of two type’s namely positive and negative \[\beta \] decay.
These are as follow: Negative \[\beta \] decay is \[{}_{\text{Z}}{{\text{X}}^{\text{A}}} \to {}_{{\text{Z}} + 1}{{\text{Y}}^{\text{A}}} + {}_ - {\beta ^0} + \bar \nu \] and Positive \[\beta \] decay is \[{}_{\text{Z}}{{\text{X}}^{\text{A}}} \to {}_{{\text{Z}} - 1}{{\text{Y}}^{\text{A}}} + {}_ + {\beta ^0} + \nu \] .
\[\gamma \] Decay: After \[\alpha \] decay or \[\beta \] decay, daughter nuclei may be in excited state and return to ground state by emitting photons of high energy called \[\gamma \] photons.
This is as follow: \[{\left( {{}_{\text{Z}}{{\text{X}}^{\text{A}}}} \right)^ * } \to {}_{\text{Z}}{{\text{X}}^{\text{A}}} + \gamma \]
Now decay of tritium can be of only \[{}_ - {\beta ^0}\] type, it is as follow: \[{}_1{{\text{T}}^3} \to {}_2{\text{H}}{{\text{e}}^3} + {{\text{e}}^ - } + \bar \nu \]

Thus, the correct option is C.
Note:
In \[\alpha \] decay, atomic number decreases by 2 and atomic mass decreases by 2. In Negative \[\beta \] decay, atomic mass is unchanged and atomic number increases by 1. In Positive \[\beta \] decay, Atomic mass is unchanged and atomic number decreases by 1. In \[\gamma \] decay, both atomic number and atomic mass remain unchanged.