Q28E - Chemistry (OpenStax)

Question

Question: Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed: 

(a) \({}_2^{\rm{6}}He\)

(b) \({}_{30}^{60}Zn\)

(c) \({}_{91}^{{\rm{235}}}Pa\)

(d) \({}_{94}^{241}{\rm{Np}}\)

(e) \({}^{18}{\rm{F}}\)

(f) \({}^{129}Ba\)

(g) \({}^{237}{\rm{Pu}}\)

Step-by-Step Solution

Answer

(a) Since \({}_2^{\rm{6}}He\) has a neutron-proton ratio of \({\rm{2}}\), it undergoes beta decay.

(b) Since \({}_{30}^{60}Zn\) has a neutron-proton ratio of \({\rm{1}}\), it undergoes positron emission.

(c) Since \({}_{91}^{235}Pa\) has a neutron-proton ratio of \({\rm{1}}{\rm{.58}}\), it undergoes alpha decay.

(d) Since \({}_{94}^{241}{\rm{Np}}\) has a neutron-proton ratio of \({\rm{1}}{\rm{.56}}\), it undergoes alpha decay.

(e) Since \({}_9^{18}{\rm{F}}\) has a neutron-proton ratio of \({\rm{1}}\), it undergoes positron emission.

(f) Since \({}_{56}^{129}Ba\) has a neutron-proton ratio of \({\rm{1}}{\rm{.30}}\), it undergoes beta decay.

(g) Since \({}_{94}^{237}{\rm{Pu}}\) has a neutron-proton ratio of \({\rm{1}}{\rm{.52}}\), it undergoes alpha decay.

Step 1: Introduction

The band of stability is the region in the graph plotted between the number of neutrons and the number of protons in which stable nuclei are present.

For the elements which have an atomic number less than \({\rm{20}}\), the ratio between neutrons and protons should be \({\rm{1}}\) for the nuclei to be stable.

For the elements which have an atomic number greater than \({\rm{20}}\) and less than \({\rm{83}}\), the ratio of neutron and proton should be \({\rm{1}}{\rm{.5}}\) for the nuclei to be stable.

• If the N/Z ratio is high, alpha decay occurs because it decreases the N/Z ratio for the elements which have an atomic number greater than 83.

• If the  N/Z  ratio is low, positron emission occurs to increase the  N/Z  ratio.

• Beta decay also decreases the  N/Z  ratio for the elements which have an atomic number less than 83 .

Step 2: Decay of \({}_2^{\rm{6}}He\)

(a)

The number of protons is \(2\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 6 - 2 = 4}}\end{array}\)

Now, calculate the ratio –

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{{\rm{4}}}{{\rm{2}}}\\{\rm{ratio = 2}}\end{array}\)

It can be seen that \({}_2^{\rm{6}}He\) has a greater N/Z  ratio.

 

Therefore, \({}_2^{\rm{6}}He\) undergoes beta decay to attain stability.

Step 3: Decay of \({}_{30}^{60}Zn\)

(b)

The number of protons is \(30\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 60 - }}30{\rm{ = }}30\end{array}\)

Now, calculate the ratio:

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{{30}}{{30}}\\{\rm{ratio = 1}}\end{array}\)

It can be seen that \({}_{30}^{60}Zn\) has a lower N/Z ratio.

 

Therefore, \({}_{30}^{60}Zn\) undergoes positron emission to attain stability.

Step 4: Decay of \({}_{91}^{{\rm{235}}}Pa\)

(c)

The number of protons is \(91\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 235 - 91 = 144}}\end{array}\)

Now, calculate the ratio:

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{{{\rm{144}}}}{{91}}\\{\rm{ratio = 1}}{\rm{.58}}\end{array}\)

It can be seen that \({}_{91}^{{\rm{235}}}Pa\) has a greater N/Z ratio.

 

Therefore, \({}_{91}^{{\rm{235}}}Pa\) undergoes alpha decay to attain stability.

Step 5:Decay of \({}_{94}^{241}{\rm{Np}}\)

(d)

The number of protons is \(94\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 241 - 94 = 147}}\end{array}\)

Now, calculate the ratio:

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{{{\rm{147}}}}{{94}}\\{\rm{ratio = 1}}{\rm{.56}}\end{array}\)

It can be seen that \({}_{94}^{241}{\rm{Np}}\) has a greater N/Z ratio.

 

Therefore, \({}_{94}^{241}{\rm{Np}}\) undergoes alpha decay to attain stability.

Step 6:Decay of \({}^{18}{\rm{F}}\)

(e)

The number of protons is \(9\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 18 - 9 = 9}}\end{array}\)

Now, calculate the ratio –

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{9}{9}\\{\rm{ratio = 1}}\end{array}\)

It can be seen that \({}_9^{18}{\rm{F}}\) has a lower N/Z ratio.

 

Therefore, \({}_9^{18}{\rm{F}}\) undergoes positron emission to attain stability.

Step 7: Decay of \({}^{129}Ba\)

(f)

The number of protons is \(56\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 129 - 56 = 73}}\end{array}\)

Now, calculate the ratio:

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{{{\rm{73}}}}{{{\rm{56}}}}\\{\rm{ratio = 1}}{\rm{.30}}\end{array}\)

It can be seen that \({}_{56}^{129}Ba\) has a greater N/Z ratio.

 

Therefore, \({}_{56}^{129}Ba\) undergoes beta decay to attain stability.

Step 8: Decay of \({}^{237}{\rm{Pu}}\)

(c)

The number of protons is \(94\), and the number of neutrons is:

\({\rm{A = N(}}{{\rm{p}}^{\rm{ + }}}{\rm{) + N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}\)

Rearranging the equation and solving:

\(\begin{array}{c}{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = A - N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}\\{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{) = 237 - 94 = 143}}\end{array}\)

Now, calculate the ratio:

\(\begin{array}{c}{\rm{ratio = }}\frac{{{\rm{N(}}{{\rm{n}}^{\rm{0}}}{\rm{)}}}}{{{\rm{N(}}{{\rm{p}}^{\rm{ + }}}{\rm{)}}}}\\{\rm{ratio = }}\frac{{{\rm{143}}}}{{94}}\\{\rm{ratio = 1}}{\rm{.52}}\end{array}\)

It can be seen that \({}_{94}^{{\rm{237}}}Pu\) has a greater N/Z ratio.

 

Therefore, \({}_{94}^{{\rm{237}}}Pu\) undergoes alpha decay to attain stability.