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

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.