Question

(a) Calculate the radius of\(^{{\rm{58}}}{\rm{Ni}}\), one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of\(^{{\rm{58}}}{\rm{Ni}}\)to that of\(^{{\rm{258}}}{\rm{Ha}}\), one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom.

Short answer

1. The radius of the \(^{{\rm{58}}}{\rm{Ni}}\) particle is obtained as: \(4.645 \times {10^{ - 15}}\,{\rm{m}}\).
2. The ratio is obtained as: \(0.608\).

Step-by-step solution

Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Evaluating the radius

a)The atomic mass number of the \(^{{\rm{58}}}{\rm{Ni}}\) particle is:

\(A = 58\)

The radius of \(^{{\rm{58}}}{\rm{Ni}}\) particle is:

\({R_o} = 1.2 \times {10^{ - 15}}\,{\rm{m}}\)

Using the following relation to obtain the value of radius as:

\(\begin{align}R &= \;{R_o}{A^{\frac{1}{3}}}\\ &= 1.2 \times {10^{ - 15}}\,m \times {(58)^{\frac{1}{3}}}\\ &= {\rm{ }}4.645 \times {10^{ - 15}}\,m\end{align}\)

Therefore, the radius is: \(4.645 \times {10^{ - 15}}\,{\rm{m}}\).

Evaluating the ratio

b) The atomic mass number of the \(^{{\rm{258}}}{\rm{Ha}}\) particle is:

\(A = 258\)

The radius of \(^{{\rm{258}}}{\rm{Ha}}\) particle is:

\({R_o} = 1.2 \times {10^{ - 15}}\,{\rm{m}}\)

Using the following relation to obtain the value of radius as:

\(\begin{align}R &= \;{R_o}{A^{\frac{1}{3}}}\\ &= 1.2 \times {10^{ - 15}}\,m \times {(258)^{\frac{1}{3}}}\\ &= {\rm{ }}7.639 \times {10^{ - 15}}\,m\end{align}\)

Dividing both the radius to obtain the ratio as:

\(\begin{align}\frac{{{R_{Ni}}}}{{{R_{Ha}}}} &= \frac{{4.645 \times {{10}^{ - 15}}\,m}}{{7.639 \times {{10}^{ - 15}}\,m}}\\ &= 0.608\end{align}\)

Therefore, the ration is: \(0.608\).