Question

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

(a) bismuth\({\rm{ - 212}}\)decays into polonium\({\rm{ - 212}}\)

(b) beryllium\({\rm{ - 8}}\)and a positron are produced by the decay of an unstable nucleus

(c) neptunium\({\rm{ - 239}}\)forms from the reaction of uranium\({\rm{ - 238}}\)with a neutron and then spontaneously converts into plutonium\({\rm{ - 239}}\)

(d) strontium\({\rm{ - 90}}\)decays into yttrium\({\rm{ - 90}}\).

Short answer

1. The balanced equation for the nuclear reaction is obtained as: \(_{{\rm{83}}}^{{\rm{212}}}{\rm{Bi}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{84}}}^{{\rm{212}}}{\rm{Po}}\).
2. The balanced equation for the nuclear reaction is obtained as:\(_{\rm{5}}^{\rm{8}}{\rm{B}} \to _{{\rm{ + 1}}}^{\rm{0}}{\rm{e + }}_{\rm{4}}^{\rm{8}}{\rm{Be}}\).
3. The balanced equation for the nuclear reaction is obtained as:\(_{{\rm{92}}}^{{\rm{238}}}{\rm{U + }}_{\rm{0}}^{\rm{1}}{\rm{n}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}}\)and\(_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{94}}}^{{\rm{239}}}{\rm{Pu}}\).
4. The balanced equation for the nuclear reaction is obtained as: \(_{{\rm{38}}}^{{\rm{90}}}{\rm{Sr}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{39}}}^{{\rm{90}}}{\rm{Y}}\).

Step-by-step solution

Step 1: Define Nuclear Chemistry

Nuclear chemistry is a branch of chemistry that studies radioactivity, nuclear processes, and atomic nuclei alterations such as nuclear transmutation and nuclear characteristics.

Writing the balanced equation

1. We have the equation:

\(_{{\rm{83}}}^{{\rm{212}}}{\rm{Bi}} \to _{\rm{y}}^{\rm{x}}{\rm{? + }}_{{\rm{84}}}^{{\rm{212}}}{\rm{Po}}\)

The mass on the left side is \({\rm{212}}\); hence we also require the total of \({\rm{212}}\) on the right:

\(\begin{array}{c}{\rm{x + 212 = 212}}\\{\rm{x = 0}}\end{array}\)

Because the atomic number on the left is \({\rm{83}}\), we must also have the total of \({\rm{83}}\) on the right:

\(\begin{array}{c}{\rm{x + 84 = 83}}\\{\rm{x = - 1}}\end{array}\)

We know that the electron is a particle with an atomic number of \({\rm{ - 1}}\) and a mass of \({\rm{0}}\);thus, the whole reaction is:

\(_{{\rm{83}}}^{{\rm{212}}}{\rm{Bi}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{84}}}^{{\rm{212}}}{\rm{Po}}\)

Therefore, the balanced equation is:

\(_{{\rm{83}}}^{{\rm{212}}}{\rm{Bi}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{84}}}^{{\rm{212}}}{\rm{Po}}\)

Writing the balanced equation

1. We have the equation:

\(_{\rm{y}}^{\rm{x}}{\rm{?}} \to _{{\rm{ + 1}}}^{\rm{0}}{\rm{e + }}_{\rm{4}}^{\rm{8}}{\rm{Be}}\)

On the right-hand side, the mass is:

\(\begin{array}{c}{\rm{0 + 8 = x}}\\{\rm{x = 8}}\end{array}\)

This implies we'll need \({\rm{8}}\) on the left side as well.

The atomic number on the right side is:

\(\begin{array}{c}{\rm{1 + 4 = 5}}\\{\rm{y = 5}}\end{array}\)

Which then means we also need \({\rm{5}}\) on the left side.

Looking at the periodic chart, the element with the atomic number \({\rm{5}}\) is B, implying that the entire reaction is:

\(_{\rm{5}}^{\rm{8}}{\rm{B}} \to _{{\rm{ + 1}}}^{\rm{0}}{\rm{e + }}_{\rm{4}}^{\rm{8}}{\rm{Be}}\)

Therefore, the balanced equation is:

\(_{\rm{5}}^{\rm{8}}{\rm{B}} \to _{{\rm{ + 1}}}^{\rm{0}}{\rm{e + }}_{\rm{4}}^{\rm{8}}{\rm{Be}}\)

Writing the balanced equation

1. We have the equation:

\(_{{\rm{92}}}^{{\rm{238}}}{\rm{U + }}_{\rm{0}}^{\rm{1}}{\rm{n}} \to _{\rm{y}}^{\rm{x}}{\rm{? + }}_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}}\)

On the left side, the total of the masses is:

\({\rm{238 + 1 = 239}}\)

This implies we'll need the total of\({\rm{239}}\)on the right side as well:

\(\begin{array}{c}{\rm{x + 239 = 239}}\\{\rm{x = 0}}\end{array}\)

The atomic number on the left adds up to:

\({\rm{92 + 0 = 92}}\)

This implies we'll also need the total of\({\rm{92}}\)on the right side:

\(\begin{array}{c}{\rm{y + 93 = 92}}\\{\rm{y = - 1}}\end{array}\)

We know that the electron is a particle with an atomic number of\({\rm{ - 1}}\)and a mass of\({\rm{0}}\);thus, the whole reaction is:

\(_{{\rm{92}}}^{{\rm{238}}}{\rm{U + }}_{\rm{0}}^{\rm{1}}{\rm{n}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}}\)

The second reaction we have is:

\(_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}} \to _{\rm{y}}^{\rm{x}}{\rm{? + }}_{{\rm{94}}}^{{\rm{239}}}{\rm{Pu}}\)

As the mass on the left side is\({\rm{239}}\), we must also have the total of\({\rm{239}}\)on the right side:

\(\begin{array}{c}{\rm{x + 239 = 239}}\\{\rm{x = 0}}\end{array}\)

The atomic number on the left is\({\rm{93}}\), which means we also need the total of\({\rm{93}}\)on the right:

\(\begin{array}{c}{\rm{y + 94 = 93}}\\{\rm{y = - 1}}\end{array}\)

We know that the electron is a particle with an atomic number of\({\rm{ - 1}}\)and a mass of\({\rm{0}}\);thus, the whole reaction is:

\(_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{94}}}^{{\rm{239}}}{\rm{Pu}}\)

Therefore, the reactions are:

\(_{{\rm{92}}}^{{\rm{238}}}{\rm{U + }}_{\rm{0}}^{\rm{1}}{\rm{n}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}}\)

\(_{{\rm{93}}}^{{\rm{239}}}{\rm{Np}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{94}}}^{{\rm{239}}}{\rm{Pu}}\)

Writing the balanced equation

1. The equation we have is:

\(_{{\rm{38}}}^{{\rm{90}}}{\rm{Sr}} \to _{\rm{y}}^{\rm{x}}{\rm{? + }}_{{\rm{39}}}^{{\rm{90}}}{\rm{Y}}\)

As the mass on the left side is\({\rm{90}}\), we must also have the total of\({\rm{90}}\)on the right side:

\(\begin{array}{c}{\rm{x + 90 = 90}}\\{\rm{x = 0}}\end{array}\)

The atomic number on the left is\({\rm{38}}\), which means we also need the total of\({\rm{38}}\)on the right:

\(\begin{array}{c}{\rm{y + 38 = 39}}\\{\rm{y = - 1}}\end{array}\)

We know that the electron is a particle with an atomic number of\({\rm{ - 1}}\)and a mass of\({\rm{0}}\);thus, the whole reaction is:

\(_{{\rm{38}}}^{{\rm{90}}}{\rm{Sr}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{39}}}^{{\rm{90}}}{\rm{Y}}\)

Therefore, the reaction is: \(_{{\rm{38}}}^{{\rm{90}}}{\rm{Sr}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{39}}}^{{\rm{90}}}{\rm{Y}}\).