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

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

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

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

(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.

(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.

(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.

(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.

(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.

(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.

(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.