Question

The equation for ²³₁₀Ne nucleus which decays by β-emission is:

• A. \[ {}^{23}_{10}\text{Ne} \xrightarrow{\beta\text{-decay}} {}^{23}_{11}\text{Na} + \overline{\nu} + e^{-} \]
• B. \[ {}^{23}_{10}\text{Ne} \xrightarrow{\beta\text{-decay}} {}^{23}_{11}\text{Na} + \overline{\nu} + e^{-} \]
• C. \[ {}^{23}_{10}\text{Ne} \xrightarrow{\beta\text{-decay}} {}^{23}_{11}\text{Na} + e^{-} + \overline{\nu} \]
• D. \[ {}^{23}_{10}\mathrm{Ne} \xrightarrow{\beta\text{-decay}} {}^{23}_{11}\mathrm{Na} + e^{-} + \overline{\nu} \]

Hint:

In \(\beta^-\)-decay, the neutron changes to a proton; atomic number increases by 1, while mass number remains constant.

Answer 1

The Correct Option is D

\(\beta^-\)-decay involves the conversion of a neutron into a proton with emission of an electron (\(e^-\)) and an antineutrino (\(\overline{\nu}\)).
So, the atomic number increases by 1, mass number stays the same.

Therefore:
\[ {}^{23}_{10}\mathrm{Ne} \xrightarrow{\beta^-} {}^{23}_{11}\mathrm{Na} + e^{-} + \overline{\nu} \]