Chapter 28, Problem 11P

(a)

To determine

The radius of the orbit.

Answer to Problem 11P

The radius of the orbit $0.212\text{nm}$.

Explanation of Solution

Given Info: The Bohr radius is $0.0529\text{nm}$.

Formula to calculate the radius is,

$r=n^2{a}_o$

• $r$ is radius of the orbit
• $n$ is energy level
• $a_o$ is the Bohr radius

Substitute $2$ for $n$, $0.0529\text{nm}$ for $a_o$ to find $r$.

$\begin{array}{c}r={\left(2\right)}^2\left(0.0529\text{nm}\right)\\ =0.212\text{nm}\end{array}$

Thus, the radius of the orbit is $0.212\text{nm}$

Conclusion:

Therefore, the radius of the orbit is $0.212\text{nm}$.

(b)

To determine

The linear momentum of the electron.

Answer to Problem 11P

The linear momentum of the electron is $9.95\times {10}^{-25}\text{kg-m/s}$.

Explanation of Solution

From unit conversion,

$1\text{nm}=1\times {10}^{-9}\text{m}$

Formula to calculate the momentum is,

$p=\sqrt{\frac{mk{e}^2}{r}}$

• $r$ is radius of the orbit
• $m$ is mass of electron
• $e$ is the charge of electron
• $k$ is Coulomb’s constant
• $p$ is momentum of the electron

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$, $0.212\text{nm}$ for $r$, $9.1\times {10}^{-31}\text{kg}$ for $m$, $9\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ to find $p$.

$\begin{array}{c}p=\sqrt{\frac{\left(9.1\times {10}^{-31}\text{kg}\right)\left(9\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{\left(0.212\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{nm}}{1\text{m}}\right)}}\\ =9.95\times {10}^{-25}\text{kg-m/s}\end{array}$

Thus, the momentum of the electron is $9.95\times {10}^{-25}\text{kg-m/s}$.

Conclusion:

Therefore, the momentum of the electron is $9.95\times {10}^{-25}\text{kg-m/s}$.

(c)

To determine

The angular momentum of the electron.

Answer to Problem 11P

The angular momentum of the electron is $2.11\times {10}^{-34}\text{J-s}$.

Explanation of Solution

Formula to calculate the angular momentum is,

$L=n\left(\frac{h}{2\pi }\right)$

• $L$ is angular momentum of the electron
• $n$ is state
• $h$ is the Planck’s constant

Substitute $2$ for $n$, $6.63\times {10}^{-34}\text{J-s}$ for $h$, to find $L$.

$\begin{array}{c}L=\left(2\right)\left(\frac{6.63\times {10}^{-34}\text{J-s}}{2\pi }\right)\\ =2.11\times {10}^{-34}\text{J-s}\end{array}$

Thus, the angular momentum of the electron is $2.11\times {10}^{-34}\text{J-s}$.

Conclusion:

Therefore, the angular momentum of the electron is $2.11\times {10}^{-34}\text{J-s}$.

(d)

To determine

The kinetic energy of the electron.

Answer to Problem 11P

The kinetic energy of the electron is $3.4\text{eV}$.

Explanation of Solution

Formula to calculate the kinetic energy is,

$K.E=\frac{p^2}{2m}$

• $p$ is linear momentum of the electron
• $m$ is mass of the electron

From unit conversion,

$1\text{eV}=1.6\times {10}^{-19}\text{J}$

Substitute $9.95\times {10}^{-25}\text{kg-m/s}$ for $p$, $9.1\times {10}^{-31}\text{kg}$ for $m$, to find $K.E$.

$\begin{array}{c}K.E=\frac{{\left(9.95\times {10}^{-25}\text{kg-m/s}\right)}^2}{2\left(9.1\times {10}^{-31}\text{kg}\right)}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =3.4\text{eV}\end{array}$

Thus, the kinetic energy of the electron is $3.4\text{eV}$.

Conclusion:

Therefore, the kinetic energy of the electron is $3.4\text{eV}$.

(e)

To determine

The potential energy of the electron.

Answer to Problem 11P

The potential energy of the electron is $-6.8\text{eV}$.

Explanation of Solution

Formula to calculate the potential energy is,

$P.E=-\frac{k{e}^2}{r}$

• $k$ is coulomb’s constant
• $e$ is the electronic charge
• $r$ is radius of the orbit

From unit conversion,

$1\text{eV}=1.6\times {10}^{-19}\text{J}$

$1\text{nm}=1\times {10}^{-9}\text{m}$

Substitute $9\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.6\times {10}^{-19}\text{C}$ for $e$, $0.212\text{nm}$ for $r$, to find $P.E$.

$\begin{array}{c}P.E=-\frac{\left(9\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{\left(0.212\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)}\left(\frac{1\text{eV}}{1\times {10}^{-19}\text{J}}\right)\\ =-6.8eV\end{array}$

Thus, the potential energy of the electron is $-6.8eV$.

Conclusion:

Therefore, the potential energy of the electron is $-6.8eV$.

(f)

To determine

The total energy of the electron.

Answer to Problem 11P

The total energy of the electron is $-3.4\text{eV}$.

Explanation of Solution

Formula to calculate the total energy is,

$E=K.E+P.E$

• $E$ is total energy
• $K.E$ is the kinetic energy electron
• $P.E$ is potential energy of the electron.

Substitute $3.4\text{eV}$ for $K.E$, $-6.8\text{eV}$ for $P.E$, to find $E$.

$\begin{array}{c}E=\left(3.4\text{eV}\right)+\left(-6.8\text{eV}\right)\\ =-3.4\text{eV}\end{array}$

Thus, the total energy of the electron is $-3.4eV$.

Conclusion:

Therefore, the total energy of the electron is $-3.4eV$