Chapter 29, Problem 8P

(a)

To determine

The orbital radii of $\text{C}$ and $\text{C}$.

Answer to Problem 8P

The orbital radii of $\text{C}$ is 7.89 cm.

The orbital radii of $\text{C}$ is 8.21 cm.

Explanation of Solution

Section 1:

To determine: The orbital radii of $\text{C}$.

Answer: The orbital radii of $\text{C}$ is 7.89 cm.

Given info: Potential (V) is $1.00\times {10}^3\text{V}$. The magnetic field strength (B) is 0.200 T.

From Law of conservation of energy,

$\Delta KE+\Delta PE=0$ (1)

• $\Delta KE$ is the kinetic energy.
• $\Delta PE$ is the potential energy.

Formula to calculate the kinetic energy is,

$\Delta KE=\frac{1}{2}m{v}^2$ (2)

• m is the mass of the atom.
• v is the velocity.

Formula to calculate the potential energy is,

$\Delta PE=-eV$ (3)

• e is the unit of charge.

Substitute Equations (2) and (3) in (1) and re-arrange to get v.

$v=\sqrt{\frac{2eV}{m}}$ (4)

Formula to calculate the radii is,

$r=\frac{mv}{eB}$ (5)

Substitute Equation (5) in (4) to get r.

$r=\sqrt{\frac{2mV}{e{B}^2}}$ (6)

Substitute 12 u for m, $1.00\times {10}^3\text{V}$ for V, 0.200 T for B and $1.6\times {10}^{-19}\text{C}$ for e in Equation (6) to get r.

$\begin{array}{c}r=\sqrt{\frac{2\left(12\text{u}\right)\left(1.00\times {10}^3\text{V}\right)}{\left(1.6\times {10}^{-19}\text{C}\right){\left(0.200\text{T}\right)}^2}}\\ =\sqrt{\frac{2\left(12\text{u}\right)\left(\frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}\right)\left(1.00\times {10}^3\text{V}\right)}{\left(1.6\times {10}^{-19}\text{C}\right){\left(0.200\text{T}\right)}^2}}\\ =7.89\text{cm}\end{array}$

The orbital radii of $\text{C}$ is 7.89 cm.

Section 2:

To determine: The orbital radii of $\text{C}$.

Answer: The orbital radii of $\text{C}$ is 8.21 cm.

Given info: Potential (V) is $1.00\times {10}^3\text{V}$. The magnetic field strength (B) is 0.200 T.

From Equation (6) of Section 1,

$r=\sqrt{\frac{2mV}{e{B}^2}}$

Substitute 13 u for m, $1.00\times {10}^3\text{V}$ for V, 0.200 T for B and $1.6\times {10}^{-19}\text{C}$ for e in the above equation to get r.

$\begin{array}{c}r=\sqrt{\frac{2\left(13\text{u}\right)\left(1.00\times {10}^3\text{V}\right)}{\left(1.6\times {10}^{-19}\text{C}\right){\left(0.200\text{T}\right)}^2}}\\ =\sqrt{\frac{2\left(13\text{u}\right)\left(\frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}\right)\left(1.00\times {10}^3\text{V}\right)}{\left(1.6\times {10}^{-19}\text{C}\right){\left(0.200\text{T}\right)}^2}}\\ =8.21\text{cm}\end{array}$

The orbital radii of $\text{C}$ is 8.21 cm.

Conclusion:

The orbital radii of $\text{C}$ is 7.89 cm.

The orbital radii of $\text{C}$ is 8.21 cm.

(b)

To determine

The relation between the radii of $\text{C}$ and $\text{C}$.

Answer to Problem 8P

The relation between the radii of $\text{C}$ and $\text{C}$ is $\frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}}$.

Explanation of Solution

Given info: Potential (V) is $1.00\times {10}^3\text{V}$. The magnetic field strength (B) is 0.200 T.

From Equation (4) of Section 1,

$r=\sqrt{\frac{2mV}{e{B}^2}}$

In the above equation, mass, potential and magnetic field are the same for the atoms. Therefore, r is proportional to $\sqrt{m}$. Therefore,

$\frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}}$ (7)

• $r_1$ and $r_2$ are the radii of $\text{C}$ and $\text{C}$ respectively.
• $m_1$ and $m_2$ are the masses of $\text{C}$ and $\text{C}$ respectively.

Substitute 12 u for $m_1$ and 13 u for $m_2$ in the above equation.

$\begin{array}{c}\frac{r_1}{r_2}=\sqrt{\frac{12\text{u}}{13\text{u}}}\\ =0.961\end{array}$ (8)

From (a), $r_1=7.89\text{cm}$ and $r_2=8.21\text{cm}$. The ratio of $r_1$ and $r_2$ is,

$\begin{array}{c}\frac{r_1}{r_2}=\frac{7.89\text{cm}}{8.21\text{cm}}\\ =0.961\end{array}$ (9)

From Equations (8) and (9), radii of $\text{C}$ and $\text{C}$ satisfy the formula in (7).

Conclusion:

The relation between the radii of $\text{C}$ and $\text{C}$ is $\frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}}$