Question: Technetium-99 is prepared from \(^{98}Mo.\) Molybdenum- 98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a \(\beta \) particle to yield an excited form of technetium-99, represented as \(^{99}T{c^*}.\) This excited nucleus relaxes to the ground state, represented as \(^{99}Tc,\) by emitting a \(\gamma \) ray. The ground state of \(^{99}Tc\) then emits a \(\beta \) particle. Write the equations for each of these nuclear reactions.

First molybdenum- 98 combines with a neutron to give molybdenum-99:

\(_{42}^{98}Mo + _0^1n \to _{42}^{99}Mo\)

Then, molybdenum-99 emits a beta particle to yield an excited form of technetium-99 (shown with\(^*\)). Beta particle is an electron and it is emmited which means it will be on the right side of our equation. Now we need to look will it be an electron or positron

\(\begin{array}{l}_{42}^{99}Mo \to _{43}^{99}T{c^*} + _y^x?\\\\42 = 43 + y\\\\y = - 1\end{array}\)

This means we have an electron:

\(_{42}^{99}Mo \to _{43}^{99}T{c^*} + _{ - 1}^0e\)

Now the excited state of technetium-99 relaxes to the ground state, by emmiting gamma ray:

\(_{43}^{99}T{c^*} \to _{43}^{99}Tc + _0^0\gamma \)

And for last, the ground state of technetium-99 emits a beta particle:

\(_{43}^{99}Tc \to _y^x? + _{ - 1}^0e\)

The mass on the left side is 99 , which means on the right side we also need to have the sum of 99 :

\(\begin{array}{l}x + 0 = 99\\x = 99\end{array}\)

The atomic number on the left is 43 , which means on the right side we also need to have the sum of 43 :

\(\begin{array}{l}y + ( - 1) = 43\\y = 44\end{array}\)

From the periodic table we can see that element which has an atomic number of 44 \(Ru\),which means the complete reaction is:

\(_{43}^{99}Tc \to _{44}^{99}Ru + _{ - 1}^0e\)