Question

The stable isotopes of Neon are ${}^{\mathbf{20}}\mathbf{Ne}$, ${}^{\mathbf{21}}\mathbf{Ne}$, and ${}^{\mathbf{22}}\mathbf{Ne}$. Predict the nuclides form when ${}^{\mathbf{19}}\mathbf{Ne}$ and ${}^{\mathbf{23}}\mathbf{Ne}$ decay.

Short answer

• ${}^{19}\text{Ne}$undergoes positron emission to yield ${}^{19}\text{F}$.
• ${}^{23}\text{Ne}$ undergoes beta emission to yield ${}^{23}\text{Na}$.

Step-by-step solution

positron emission

Number of protons in ${}^{19}\text{Ne} =10$

Number of neutrons in ${}^{19}\text{Ne} =9$

The number of neutrons is less than the number of protons. Thus, we can predict that a positron emission is possible. Let us write down the balanced nuclear reaction by ${}^{\text{19}}\text{Ne}$.

${}_{10}{}^{19}Ne\to {}_9{}^{19}F+{}_1{}^0{e}^{+}+\nu$

To confirm that the above reaction is feasible or not, we have to calculate the mass difference.

$$\begin{gathered} \Delta m=m\left({}_9{}^{19}F\right)+2m\left({}_{+1}{}^0{e}^{+}\right)-m\left({}_{10}{}^{19}Ne\right) \\ =18.9984u+2\times 0.0005485u-20.1797 u \\ =-1.180 u \end{gathered}$$

Since $\Delta m<0$, we can predict that the positron emission would be spontaneous and ${}_9{}^{19}\text{F}$would be formed from the positron emission

β  emission

Number of protons in ${}^{23}\text{Ne} =10$

Number of neutrons in ${}^{23}\text{Ne} =13$

The number of neutrons is greater than the number of protons, so we can predict that an electron emission is possible. Let us write down the balanced nuclear reaction for electron emission by ${}^{23}\text{F}$.

${}_{10}{}^{23}Ne\to {}_{11}{}^{23}Na+{}_{-1}{}^0{e}^{-}+\overline{\nu }$

We have to calculate the mass difference to predict the spontaneity of the electron emission.

$$\begin{gathered} \Delta m=m\left({}_{11}{}^{23}Na\right)-m\left({}_{10}{}^{23}Ne\right) \\ =22.9897u-23.0000u \\ =-0.0103u \end{gathered}$$

Since $\Delta m<0$, we can predict that the electron emission would be spontaneous and ${}_{10}{}^{23}\text{Na}$ would be formed from the electron emission.