Question:15. Write a balanced equation for each of the following nuclear reactions:

1. The product of \(^{{\bf{17}}}{\bf{O}}\) from \(^{{\bf{14}}}{\bf{N}}\) by \({\bf{\alpha }}\) particle bombardment
2. The production of \(^{{\bf{14}}}{\bf{C}}\) from \(^{{\bf{14}}}{\bf{N}}\) by neutron bombardment
3. The production of \(^{{\bf{233}}}{\bf{Th}}\) from \(^{{\bf{232}}}{\bf{Th}}\) by neutron bombardment
4. The production of \(^{{\bf{239}}}{\bf{U}}\) from \(^{{\bf{238}}}{\bf{U}}\) by \({_{\bf{1}}^{\bf{2}}}{\bf{H}}\) bombardment\(\)

We have:\(_7^{14}\;N + _2^4He \to _8^{17}O\)

We can see that this equation is not balanced and that we are missing one product :

\(_7^{14}\;N + _2^4He \to _8^{17}O + _y^x?\)

The sum of the masses on the left side is:

\(14 + 4 = 18\)

Which means on the right side we also need to have the sum of 18 :

\(\begin{array}{l}x + 17 = 18\\x = 1\end{array}\)

The sum of the atomic number on the left is:

\(7 + 2 = 9\)

Which means on the right side we also need to have the sum of 9:

\(\begin{array}{l}y + 8 = 9\\y = 1\end{array}\)

From the periodic table we can see that element which has an atomic number of 1 is\(H,\)which means the complete reaction is:

\(_7^{14}\;N + _2^4He \to _8^{17}O + _1^1H\)

We have\(_7^{14}N + _0^1n \to _6^{14}C\)

We can see that this equation is not balanced and that we are missing one product:

\(_7^{14}N + _0^1n \to _6^{14}C + _y^x?\)

The sum of the masses on the left side is:

\(14 + 1 = 15\)

Which means on the right side we also need to have the sum of 15:

\(\begin{array}{l}x + 14 = 15\\x = 1\end{array}\)

The sum of the atomic number on the left is:

\(7 + 0 = 7\)

Which means on the right side we also need to have the sum of 7 :

\(\begin{array}{l}y + 6 = 7\\y = 1\end{array}\)

From the periodic table we can see that element which has an atomic number of 1 is H, which means the complete reaction is:

\(_7^{14}N + \frac{1}{0}n \to _6^{14}C + _1^1H\)

We have:\(_{90}^{232}Th + \frac{1}{0}n \to 9_{90}^{233}Th.\)

We can see that this equation is balanced

We have:\(_{92}^{238}U + _1^2H \to _{92}^{239}U\)

We can see that this equation is not balanced and that we are missing one product:

\(_{92}^{238}U + _1^2H \to _{92}^{239}U + _y^x?\)

The sum of the masses on the left side is:

\(238 + 2 = 240\)

Which means on the right side we also need to have the sum of 240 :

\(\begin{array}{l}x + 239 = 240\\x = 1\end{array}\)

The sum of the atomic number on the left is:

\(92 + 1 = 93\)

Which means on the right side we also need to have the sum of 93 :

\(\begin{array}{l}y + 92 = 93\\y = 1\end{array}\)

From the periodic table we can see that element which has an atomic number of 1 is H, which means the complete reaction is

\(_{92}^{238}U + _1^2H \to _{92}^{239}U + _1^1H\)