Chapter 27, Problem 37QAP

To determine

(A)

The radius of hydrogen-3 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{hydrogen-3:}\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(\text{3}\right)}^{\frac{1}{3}}\\ r=\text{1}\text{.7 fm}\end{array}$

Thus, the radius of hydrogen-3 of the nuclei of the isotopes is $r=\text{1}\text{.7 fm }$.

To determine

(B)

The radius of beryllium-8 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{beryllium-8: }\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(8\right)}^{\frac{1}{3}}\\ r\text{= 2}\text{.4 fm}\end{array}$

Thus, the radius of beryllium-8 of the nuclei of the isotopes is $r\text{=}\text{2}\text{.4 fm }$.

To determine

(C)

The radius of aluminum-26 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{aluminum-26:}\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(26\right)}^{\frac{1}{3}}\\ r\text{=3}\text{.6 fm }\end{array}$

Thus, the radius of aluminum-26 of the nuclei of the isotopes is $r\text{=3}\text{.6 fm }$.

To determine

(D)

The radius of gold-197 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{gold-197:}\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(197\right)}^{\frac{1}{3}}\\ r\text{=7}\text{.0 fm}\end{array}$

Thus, the radius of gold-197 of the nuclei of the isotopes is $r\text{=7}\text{.0 fm }$.

To determine

(E)

The radius of technetium-100 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{technetium-100:}\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(100\right)}^{\frac{1}{3}}\\ r\text{=5}\text{.6 fm }\end{array}$

Thus, the radius of technetium-100 of the nuclei of the isotopes is $r\text{=5}\text{.6 fm }$.

To determine

(F)

The radius of tungsten-184 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{tungsten-184: }\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(184\right)}^{\frac{1}{3}}\\ r\text{=6}\text{.8 fm}\end{array}$

Thus, the radius of tungsten-184 of the nuclei of the isotopes is $r\text{=6}\text{.8 fm }$.

To determine

(G)

The radius of osmium-190 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{osmium-190:}\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(190\right)}^{\frac{1}{3}}\\ r=\text{6}\text{.9 fm }\end{array}$

Thus, the radius of osmium-190 of the nuclei of the isotopes is $r=\text{6}\text{.9 fm}$.

To determine

(H)

The radius of plutonium-239 of the nuclei of the isotope.

Explanation of Solution

An atomic nucleus is approximately spherical with radius $r=r_o{A}^{\frac{1}{3}}$, where $r_0\text{ = }1.2\times {10}^{-15}\text{m}$ and A is the mass number.

  $\begin{array}{l}\text{plutonium-239:}\\ r=r_o{A}^{\frac{1}{3}}\\ r=\left(\text{1}\text{.2 fm}\right){\left(239\right)}^{\frac{1}{3}}\\ r=\text{7}\text{.4 fm}\end{array}$

Thus, the radius of plutonium-239 of the nuclei of the isotopes is $r\text{=}\text{7}\text{.4 fm }$.