Question

Question: The natural abundance of ${}^{\mathbf{30}}\mathbf{Si}$ is 3.1%. Upon irradiation with neutrons, the isotope is converted to${}^{\mathbf{31}}\mathbf{Si}$ , which decays to the stable isotope ${}^{\mathbf{31}}\mathbf{P}$. This provides a way for introducing trace amounts of phosphorus into silicon in a much more uniform fashion then possible by ordinary mixing of silicon and phosphorus and gives semiconductor devices the capability of handling much higher levels of power. Write the nuclear reaction for the two steps in the preparation of ${}^{\mathbf{31}}\mathbf{P}$from ${}^{\mathbf{30}}\mathbf{Si}$.

Short answer

The balanced nuclear reaction representing the formation of ${}^{31}\mathrm{P}$from ${}^{31}\mathrm{Si}$ is:

$$${}_{14}{}^{30}\mathrm{Si}+{}_0{}^1\mathrm{n}\to {}_{14}{}^{31}\mathrm{Si}\to {}_{15}{}^{31}\mathrm{P}+{}_{-1}{}^0\mathrm{e}^{-}+\overline{\mathrm{\nu }}$

Step-by-step solution

Nuclear reaction representing formation of Si31

In the first step of the reaction ${}^{30}\mathrm{Si}$ is bombarded with neutrons. The product formed is ${}^{31}\mathrm{Si}$. The balanced nuclear reaction representing the change is:

${}^{30}\mathrm{Si}+{}_0{}^1\mathrm{n}\to {}^{31}\mathrm{Si}$

Nuclear reaction representing formation of P31

Number of protons in ${}^{31}\mathrm{Si}=14$

Number of neutrons in ${}^{31}\mathrm{Si}=17$

Since the number of neutrons are more than the number of protons it can be said formation of ${}^{31}\mathrm{P}$can take place through electron emission. The balanced nuclear reaction representing electron emission by ${}^{31}\mathrm{Si}$and formation of ${}^{31}\mathrm{P}$is:

${}_{14}{}^{31}\mathrm{Si}\to {}_{15}{}^{31}\mathrm{P}+{}_{-1}{}^0\mathrm{e}^{-}+\overline{\mathrm{\nu }}$