which of the following nuclei lie within the band of stability shown in Figure\(21.2\)?

(a) chlorine-37

(b) calcium-40

(c)\(^{204}Bi\)

(d)\(^{56}Fe\)

(e)\(^{206}\;Pb\)

(f)\(^{211}\;Pb\)

(g)\(^{222}Rn\)

(h) carbon-14

Band of stability is the region in the graph plotted between the number of neutrons and number of protons in which stable nucleus are present.

We can look at the Figure \(21.2\) from which we can see the number of neutrons and protons. The stable nuclides are indicated in blue, and the unstable nuclides are indicated in green. But we can also check our result by calculating ratio of number of neutrons and number of protons.

For the elements which have the atomic number less than 20, the ratio between neutrons and protons should be 1 for the nuclei to be stable.

For the elements which have atomic number greater than 20 and less than 83, the ratio of neutron and proton should be \(1.5\)for the nuclei to be stable.

We have chlorine-37.

If we look at the periodic table the atomic number of\(Cl\)is 17, which means the number of protons is\(17.\)

Now we need to get the number of neutrons:

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

Which means:

\(\begin{aligned}{}N\left( {{n^0}} \right) &= A - N\left( {{p^ + }} \right)\\N\left( {{n^0}} \right) &= 37 - 17 &= 20\end{aligned}\)

Now we can calculate the ratio:

\(\begin{aligned}{}{\rm{ratio}} &= \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\\{\rm{ratio}} &= \frac{{20}}{{17}}{\rm{ }}\\{\rm{ratio}} &= 1.17 \approx 1\end{aligned}\)

From that we can see that chlorine-37 is stable.

We have calcium-40.

If we look at the periodic table the atomic number of\({\rm{Ca}}\)is 20, which means the number of protons is 20. Now we need to get the number of neutrons:

\(\begin{aligned}{}N\left( {{n^0}} \right) &= A - N\left( {{p^ + }} \right)\\N\left( {{n^0}} \right) &= 40 - 20 &= 20\end{aligned}\)

Now we can calculate the ratio:

\(\begin{aligned}{}{\rm{ratio }} &= \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\\{\rm{ratio }} &= \frac{{20}}{{20}}\\{\rm{ratio }} &= 1\end{aligned}\)

From that we can see that calcium-40 is stable.

Given bismuth-204.

If we look at the periodic table the atomic number of Bi is 83, which means the number of protons is\(83.\)

Now we need to get the number of neutrons:

\(\begin{aligned}{}N\left( {{n^0}} \right) &= A - N\left( {{p^ + }} \right)\\N\left( {{n^0}} \right) = 204 - 83 &= 121\end{aligned}\)

Now we can calculate the ratio:

\(\begin{aligned}{}{\rm{ratio }} &= \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\\{\rm{ratio }} &= \frac{{121}}{{83}}\\{\rm{ratio }} &= 1.46 \approx 1.5\end{aligned}\)

From that we can see that bismuth-204 is stable.

Given iron-56.

If we look at the periodic table the atomic number of Fe is 26, which means the number of protons is\(26.\)

Now we need to get the number of neutrons:

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

Which means:

\(\begin{aligned}{}N\left( {{n^0}} \right) &= A - N\left( {{p^ + }} \right)\\N\left( {{n^0}} \right) &= 56 - 26 &= 30\end{aligned}\)

Now we can calculate the ratio:

\(\begin{aligned}{}{\rm{ratio }} &= \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\\{\rm{ratio }} &= \frac{{30}}{{26}}\\{\rm{ratio }} &= 1.15 \approx 1\end{aligned}\)

From that we can see that iron-56 is stable.

Given lead-206.

If we look at the periodic table the atomic number of\({\rm{Pb}}\)is 82, which means the number of protons is\(82.\)

Now we need to get the number of neutrons:

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

Which means:

\(\begin{aligned}{}N\left( {{n^0}} \right) &= A - N\left( {{p^ + }} \right)\\N\left( {{n^0}} \right) = 206 - 82 &= 124\end{aligned}\)

Now we can calculate the ratio:

\(\begin{aligned}{}{\rm{ratio }} &= \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\\{\rm{ratio }} &= \frac{{124}}{{82}}\\{\rm{ratio }} &= 1.51 \approx 1.5\end{aligned}\)

From that we can see that lead-206 is stable.

Given lead-211.

If we look at the periodic table the atomic number of\({\rm{Pb}}\)is 82, which means the number of protons is\(82.\)now we need to get the number of neutrons:

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

Which means:

\(N\left( {{n^0}} \right) = A - N\left( {{p^ + }} \right)\)

\(N\left( {{n^0}} \right) = 211 - 82 = 129\)

Now we can calculate the ratio:

\(\begin{aligned}{}{\rm{ratio}} &= \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\\{\rm{ratio}} &= \frac{{129}}{{82}}{\rm{ }}\\{\rm{ratio}} &= 1.6 \approx 2\end{aligned}\)

From that we can see that lead-2ll is not stable.

Given radon-222.

If we look at the periodic table the atomic number of \({\rm{Rn}}\)is 86. It is known that nuclei which has atomic number greater than 83 is unstable in nature, which means radon-222 is not stable.

Given carbon-14.

If we look at the periodic table the atomic number of\(C\)is 6, which means the number of protons is 6.

Now we need to get the number of neutrons:

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

Which means:

\(\begin{aligned}{}N\left( {{n^0}} \right) &= A - N\left( {{p^ + }} \right)\\N\left( {{n^0}} \right) &= 14 - 6 &= 8\end{aligned}\)

Now we can calculate the ratio:

ratio\( = \frac{{N\left( {{n^0}} \right)}}{{N\left( {{p^ + }} \right)}}\)

ratio\( = \frac{8}{6}\)

ratio\( = 1.33 \approx 1.5\)