Q9 PE - College Physics (Urone)

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.

Step-by-Step Solution

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 1: 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.

Step 2: 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: \(4.645 \times {10^{ - 15}}\,{\rm{m}}\).

Step 2: 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: \(0.608\).