Question: Write a nuclear reaction for each step in the formation of \({}_{84}^{{\rm{218}}}Po\) from \({}_{{\rm{98}}}^{{\rm{238}}}{\rm{U}}\), which proceeds by a series of decay reactions involving the step-wise emission of \({\rm{\alpha , \beta , \beta , \alpha , \alpha , \alpha }}\) particles, in that order.

Alpha decay, also known as\({\rm{\alpha - }}\)decay, is a type of radioactive decay in which an atomic nucleus produces an alpha particle (helium nucleus) and converts or 'decays' into a new atomic nucleus with a mass number of four and an atomic number of two.

In nuclear physics, beta decay (\({\rm{\beta - }}\)decay) is a type of radioactive decay in which an atomic nucleus emits a beta particle (fast, energetic electron or positron) that transforms the original nuclide into an isobar of that nuclide.

Firstly, there is alpha emission\({\rm{(}}{}_{\rm{2}}^{\rm{4}}{\rm{He)}}\):

\(_{{\rm{98}}}^{{\rm{238}}}{\rm{U}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{238}}\), which means on the right side, there should be the sum of\({\rm{238}}\):

\(\begin{array}{c}{\rm{x + 4 = 238}}\\{\rm{x = 234}}\end{array}\)

The atomic number on the left is\({\rm{92}}\), which means on the right side, there should be the sum of\({\rm{92}}\):

\(\begin{array}{c}{\rm{y + 2 = 92}}\\{\rm{y = 90}}\end{array}\)

From the periodic table, it can be seen that element that has an atomic number of\({\rm{90}}\)is\({\rm{Th}}\), which means the complete reaction is :

\(_{{\rm{98}}}^{{\rm{238}}}{\rm{U}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{{\rm{90}}}^{{\rm{234}}}{\rm{Th}}\)

Therefore, the reaction is \(_{{\rm{98}}}^{{\rm{238}}}{\rm{U}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{{\rm{90}}}^{{\rm{234}}}{\rm{Th}}\).

There is beta emission\({\rm{(}}{}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e)}}\):

\(_{{\rm{90}}}^{{\rm{234}}}{\rm{Th}} \to {}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{234}}\), which means on the right side, there should be the sum of\({\rm{234}}\):

\(\begin{array}{c}{\rm{x + 0 = 234}}\\{\rm{x = 234}}\end{array}\)

The atomic number on the left is\({\rm{90}}\), which means on the right side, there should be the sum of\({\rm{90}}\):

\(\begin{array}{c}{\rm{y + ( - 1) = 90}}\\{\rm{y = 91}}\end{array}\)

From the periodic table, it can be seen that element that has an atomic number of\({\rm{91}}\)is\({\rm{Pa}}\), which means the complete reaction is:

\(_{{\rm{90}}}^{{\rm{234}}}{\rm{Th}} \to {}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}{}_{91}^{234}{\rm{Pa}}\)

Therefore, the reaction is \(_{{\rm{90}}}^{{\rm{234}}}{\rm{Th}} \to {}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}{}_{91}^{234}{\rm{Pa}}\).

There is beta emission\({\rm{(}}{}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e)}}\)–

\(_{{\rm{91}}}^{{\rm{234}}}{\rm{Pa}} \to {}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{234}}\), which means on the right side, there should be the sum of\({\rm{234}}\):

\(\begin{array}{c}{\rm{x + 0 = 234}}\\{\rm{x = 234}}\end{array}\)

The atomic number on the left is\({\rm{91}}\), which means on the right side, there should be the sum of\({\rm{91}}\):

\(\begin{array}{c}{\rm{y + ( - 1) = 91}}\\{\rm{y = 92}}\end{array}\)

From the periodic table, it can be seen that element that has an atomic number of\({\rm{92}}\)is\({\rm{U}}\), which means the complete reaction is:

\({}_{91}^{234}{\rm{Pa}} \to {}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}{}_{92}^{234}{\rm{U}}\)

Therefore, the reaction is \({}_{91}^{234}{\rm{Pa}} \to {}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}{}_{92}^{234}{\rm{U}}\).

There is alpha emission\({\rm{(}}{}_{\rm{2}}^{\rm{4}}{\rm{He)}}\):

\(_{{\rm{92}}}^{{\rm{234}}}{\rm{U}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{234}}\), which means on the right side, there should be the sum of\({\rm{234}}\):

\(\begin{array}{c}{\rm{x + 4 = 234}}\\{\rm{x = 230}}\end{array}\)

The atomic number on the left is\9{\rm{92}}\), which means on the right side, there should be the sum of\({\rm{92}}\):

\(\begin{array}{c}{\rm{y + 2 = 92}}\\{\rm{y = 90}}\end{array}\)

From the periodic table, it can be seen that element that has an atomic number of\({\rm{90}}\)is\({\rm{Th}}\), which means the complete reaction is:

\(_{{\rm{92}}}^{{\rm{234}}}{\rm{U}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{{\rm{90}}}^{{\rm{230}}}{\rm{Th}}\)

Therefore, the reaction is \(_{{\rm{92}}}^{{\rm{234}}}{\rm{U}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{{\rm{90}}}^{{\rm{230}}}{\rm{Th}}\).

There is alpha emission\({\rm{(}}{}_{\rm{2}}^{\rm{4}}{\rm{He)}}\):

\(_{{\rm{90}}}^{{\rm{230}}}{\rm{Th}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{230}}\), which means on the right side, there should be the sum of\({\rm{230}}\):

\(\begin{array}{c}{\rm{x + 4 = 230}}\\{\rm{x = 226}}\end{array}\)

The atomic number on the left is\({\rm{90}}\), which means on the right side, there should be the sum of\({\rm{90}}\):

\(\begin{array}{c}{\rm{y + 2 = 90}}\\{\rm{y = 88}}\end{array}\)

From the periodic table, it can be seen that element that has an atomic number of\({\rm{88}}\)is\({\rm{Ra}}\), which means the complete reaction is:

\(_{{\rm{90}}}^{{\rm{230}}}{\rm{Th}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}{}_{88}^{226}{\rm{Ra}}\)

Therefore, the reaction is \(_{{\rm{90}}}^{{\rm{230}}}{\rm{Th}} \to _{\rm{2}}^{\rm{4}}{\rm{He + }}{}_{88}^{226}{\rm{Ra}}\).

There is alpha emission\({\rm{(}}{}_{\rm{2}}^{\rm{4}}{\rm{He)}}\):

\(_{88}^{{\rm{226}}}Ra \to _{\rm{2}}^{\rm{4}}{\rm{He + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{226}}\), which means on the right side, there should be the sum of\({\rm{226}}\):

\(\begin{array}{c}{\rm{x + 4 = 226}}\\{\rm{x = 222}}\end{array}\)

The atomic number on the left is\({\rm{88}}\), which means on the right side, there should be the sum of\({\rm{88}}\):

\(\begin{array}{c}{\rm{y + 2 = 88}}\\{\rm{y = 86}}\end{array}\)

From the periodic table, it can be seen that element that has an atomic number of\({\rm{86}}\)is\({\rm{Rn}}\), which means the complete reaction is:

\(_{88}^{{\rm{226}}}Ra \to _{\rm{2}}^{\rm{4}}{\rm{He + }}{}_{86}^{222}{\rm{Rn}}\)

Therefore, the reaction is \(_{88}^{{\rm{226}}}Ra \to _{\rm{2}}^{\rm{4}}{\rm{He + }}{}_{86}^{222}{\rm{Rn}}\).

There is\({}_{86}^{222}{\rm{Rn}}\),and the product is\({}_{84}^{218}{\rm{Po}}\)–

\(_{86}^{{\rm{222}}}Rn \to {}_{84}^{218}{\rm{Po + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is\({\rm{222}}\), which means on the right side, there should be the sum of\({\rm{222}}\):

\(\begin{array}{c}{\rm{218 + x = 222}}\\{\rm{x = 4}}\end{array}\)

The atomic number on the left is\({\rm{86}}\), which means on the right side, there should be the sum of\({\rm{86}}\):

\(\begin{array}{c}{\rm{y + 84 = 86}}\\{\rm{y = 2}}\end{array}\)

This means there is alpha emission once more. So the complete reaction is –

\(_{86}^{{\rm{222}}}Rn \to {}_2^4He{\rm{ + }}{}_{{\rm{84}}}^{{\rm{218}}}{\rm{Po}}\)

Therefore, the reaction is \(_{86}^{{\rm{222}}}Rn \to {}_2^4He{\rm{ + }}{}_{{\rm{84}}}^{{\rm{218}}}{\rm{Po}}\).