Chapter 11, Problem 4C

To determine

(a)

The size of aluminum-27 and compare it with the size of first Bohr orbit in hydrogen atom.

Answer to Problem 4C

The size of aluminum-27 is compare it with the size of first Bohr orbit in hydrogen atom.

Explanation of Solution

Given info:

• The size of first Bohr orbit in hydrogen atom is $5.29\times {10}^{-11}\text{m}$.
• The Empirical formula is given to find the size of isotope.

Formula used:

Formula to find the size of isotope using formula of radius is,

$\begin{array}{c}V\text{=}\frac{4}{3}\pi r^3\\ V=\frac{4}{3}\times 3.14\times {\left(1.2\times {10}^{-15}\times A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^3\\ V=7.238\times {10}^{-45}\times A\end{array}$

Here, $A$ is the atomic mass, $V$ is the volume of nucleus.

Calculation:

Substitute the given values to find the size of aluminum-27.

$\begin{array}{c}V=7.238\times {10}^{-45}\times 39.96238\\ =2.892\times {10}^{-43}{m}^3\end{array}$

Finding the size of first Bohr orbit:

$\begin{array}{c}V\text{'}\text{=}\frac{4}{3}\pi r{\text{'}}^3\\ V=\frac{4}{3}\times 3.14\times {\left(5.29\times {10}^{-11}\text{m}\right)}^3\\ V=6.2\times {10}^{-31}{m}^3\end{array}$

The size of first Bohr orbit in hydrogen atom is $\text{3}\text{.175}\times {\text{10}}^{12}$ times more than the size of aluminum-27.

Conclusion:

Thus, the size of first Bohr orbit in hydrogen atom is $\text{3}\text{.175}\times {\text{10}}^{12}$ times more than the size of aluminum-27.

To determine

(c)

To find the size of copper-63 and compare it with the size of first Bohr orbit in hydrogen atom.

Answer to Problem 4C

The size of copper-63 is compare it with the size of first Bohr orbit in hydrogen atom.

Explanation of Solution

Given info:

• The size of first Bohr orbit in hydrogen atom is $5.29\times {10}^{-11}\text{m}$.
• The Empirical formula is given to find the size of isotope.

Formula used:

Formula to find the size of isotope using formula of radius is,

$\begin{array}{c}V\text{=}\frac{4}{3}\pi r^3\\ V=\frac{4}{3}\times 3.14\times {\left(1.2\times {10}^{-15}\times A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^3\\ V=7.238\times {10}^{-45}\times A\end{array}$

Here,

$A$ is the atomic mass, $V$ is the volume of nucleus.

Calculation:

Substitute the given values to find the size of copper-63.

$\begin{array}{c}V=7.238\times {10}^{-45}\times 62.92959\\ =4.55\times {10}^{-43}{m}^3\end{array}$

Finding the size of first Bohr orbit:

$\begin{array}{c}V\text{'}\text{=}\frac{4}{3}\pi r{\text{'}}^3\\ V=\frac{4}{3}\times 3.14\times {\left(5.29\times {10}^{-11}\text{m}\right)}^3\\ V=6.2\times {10}^{-31}{m}^3\end{array}$

The size of first Bohr orbit in hydrogen atom is $\text{1}\text{.36}\times {\text{10}}^{12}$ times more than the size of argon-40.

Conclusion:

Thus, the size of first Bohr orbit in hydrogen atom is $\text{2}\text{.1435}\times {\text{10}}^{12}$ times more than the size of copper-63.

To determine

(d)

To find the size of cesium-134 and compare it with the size of first Bohr orbit in hydrogen atom.

Answer to Problem 4C

The size of cesium-134 is compare it with the size of first Bohr orbit in hydrogen atom.

Explanation of Solution

Given info:

• The size of first Bohr orbit in hydrogen atom is $5.29\times {10}^{-11}\text{m}$.
• The Empirical formula is given to find the size of isotope.

Formula used:

Formula to find the size of isotope using formula of radius is,

$\begin{array}{c}V\text{=}\frac{4}{3}\pi r^3\\ V=\frac{4}{3}\times 3.14\times {\left(1.2\times {10}^{-15}\times A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^3\\ V=7.238\times {10}^{-45}\times A\end{array}$

Here, $A$ is the atomic mass, $V$ is the volume of nucleus.

Calculation:

Substitute the given values to find the size of cesium-134.

$\begin{array}{c}V=7.238\times {10}^{-45}\times 133.9067\\ =9.691\times {10}^{-43}{m}^3\end{array}$

Finding the size of first Bohr orbit:

$\begin{array}{c}V\text{'}\text{=}\frac{4}{3}\pi r{\text{'}}^3\\ V=\frac{4}{3}\times 3.14\times {\left(5.29\times {10}^{-11}\text{m}\right)}^3\\ V=6.2\times {10}^{-31}{m}^3\end{array}$

The size of first Bohr orbit in hydrogen atom is $\text{6}\text{.4}\times {\text{10}}^{11}$ times more than the size of cesium-134.

Conclusion:

Thus, the size of first Bohr orbit in hydrogen atom is $\text{6}\text{.4}\times {\text{10}}^{11}$ times more than the size of cesium-134.

To determine

(d)

The size of bismuth-211 and compare it with the size of first Bohr orbit in hydrogen atom.

Answer to Problem 4C

The size of bismuth-211 is compare it with the size of first Bohr orbit in hydrogen atom.

Explanation of Solution

Given info:

• The size of first Bohr orbit in hydrogen atom is $5.29\times {10}^{-11}\text{m}$
• The Empirical formula is given to find the size of isotope.

Formula used:

Formula to find the size of isotope using formula of radius is,

$\begin{array}{c}V\text{=}\frac{4}{3}\pi r^3\\ V=\frac{4}{3}\times 3.14\times {\left(1.2\times {10}^{-15}\times A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^3\\ V=7.238\times {10}^{-45}\times A\end{array}$

Here, $A$ is the atomic mass, $V$ is the volume of nucleus.

Calculation:

Substitute the given values to find the size of bismuth-211.

$\begin{array}{c}V=7.238\times {10}^{-45}\times 210.98726\\ =1.527\times {10}^{-42}{m}^3\end{array}$

Finding the size of first Bohr orbit:

$\begin{array}{c}V\text{'}\text{=}\frac{4}{3}\pi r{\text{'}}^3\\ V=\frac{4}{3}\times 3.14\times {\left(5.29\times {10}^{-11}\text{m}\right)}^3\\ V=6.2\times {10}^{-31}{m}^3\end{array}$

The size of first Bohr orbit in hydrogen atom is $\text{4}\text{.06}\times {\text{10}}^{11}$ times more than the size of bismuth-211.

Conclusion:

Thus, the size of first Bohr orbit in hydrogen atom is $\text{4}\text{.06}\times {\text{10}}^{11}$ times more than the size of bismuth-211.