Chapter 42, Problem 18P

(a)

To determine

The radius of the orbit.

Answer to Problem 18P

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

Explanation of Solution

Write the expression for the radius of any orbit in the hydrogen atom.

    $r_n=n^2{a}_0$                                                                                                                  (I)

Here, $r_n$ is the radius of the $n^{th}$ orbit in the hydrogen atom, $n$ is the order, $a_0$ is the Bohr radius.

Conclusion:

Substitute $2$ for $n$, $0.0529\text{nm}$ for $a_0$ in equation (I) to find $r_n$.

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

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

(b)

To determine

The linear momentum of the electron.

Answer to Problem 18P

The linear momentum of the electron is $\underset{\_}{9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}}$.

Explanation of Solution

The condition for the quantization of angular momentum says that for an circular orbit,

    $mvr=n\hslash$                                                                                                   (II)

Here, $m$ is the mass of the electron, $v$ is the velocity of the particle, $r$ is the radius of the orbit, $\hslash$ is equal to $\frac{1}{2\pi }$ times the Planck’s constant.

Write the expression for the linear momentum of the electron.

    $p=mv$                                                                                                     (III)

Here, $p$ is the linear momentum, $m$ is the mass of the electron, $v$ is the velocity of the electron.

Use equation (II) in equation (III), to find $p$.

    $\begin{array}{c}mv=\frac{n\hslash }{r}\\ =\frac{nh}{2\pi r}\end{array}$                                                                                                          (IV)

Conclusion:

Substitute $2$ for $n$ , $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $0.212\text{nm}$ for $r$ in equation (IV) to find $p$.

    $\begin{array}{c}p=\frac{2\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2\pi \left(0.212\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}\\ =9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Therefore, the linear momentum of the electron is $\underset{\_}{9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}}$.

(c)

To determine

The angular momentum of the electron.

Answer to Problem 18P

The angular momentum of the electron is $\underset{\_}{2.11\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}}$

Explanation of Solution

Write the expression for the angular momentum of the electron.

    $L=mvr$                                                                                                                  (V)

Conclusion:

Substitute $9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}$ for $mv$ and $0.212\text{nm}$ for $r$ in equation (V) to find $L$.

    $\begin{array}{c}L=\left(9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}\right)\left(0.212\text{nm}\right)\\ =\left(9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}\right)\left(0.212\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)\\ =2.11\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Therefore, the angular momentum of the electron is $\underset{\_}{2.11\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}}$

(d)

To determine

The kinetic energy of the electron.

Answer to Problem 18P

The kinetic energy of the electron is $\underset{\_}{3.40\text{eV}}$.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

    $K=\frac{1}{2}{m}_e{v}^2$                                                                                                           (VI)

Rearrange equation (III) to find the velocity of electron.

    $v=\frac{p}{m_e}$                                                                                                                  (VII)

Conclusion:

Substitute $9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}$ for $p$ and $9.11\times {10}^{-31}\text{kg}$ for $m_e$ in equation (VII) to find $v$.

    $\begin{array}{c}v=\frac{9.97\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}}{9.11\times {10}^{-31}\text{kg}}\\ =1.09\times {10}^6\text{m}/\text{s}\end{array}$

Substitute $1.09\times {10}^6\text{m}/\text{s}$ for $v$ and $9.11\times {10}^{-31}\text{kg}$ for $m_e$ in equation (VI) to find $K$.

  $\begin{array}{c}K=\frac{1}{2}\left(9.11\times {10}^{-31}\text{kg}\right){\left(1.09\times {10}^6\text{m}/\text{s}\right)}^2\\ =5.45\times {10}^{-19}\text{J}\times \frac{1\text{eV}}{1.602\times {10}^{-19}\text{J}}\\ =3.40\text{eV}\end{array}$

Therefore, the kinetic energy of the electron is $\underset{\_}{3.40\text{eV}}$.

(e)

To determine

The potential energy of the system.

Answer to Problem 18P

The potential energy of the system is $\underset{\_}{-6.80\text{eV}}$.

Explanation of Solution

Write the expression for the potential energy.

    $U=-\frac{k{e}^2}{r}$                                                                                                           (VIII)

Here, $U$ is the potential energy, $k$ is the constant, $e$ is the charge of electron.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $1.602\times {10}^{-19}\text{C}$ for $e$ and $0.212\text{nm}$ for $r$ in equation (VIII) to find $U$.

    $\begin{array}{c}U=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2}{0.212\text{nm}}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2}{0.212\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{nm}}}\\ =-1.09\times {10}^{-18}\text{J}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\\ \text{=}-\text{6}\text{.80}\text{eV}\end{array}$

Therefore, the potential energy of the electron is $\underset{\_}{-6.80\text{eV}}$.

(f)

To determine

The total energy of the system.

Answer to Problem 18P

The total energy of the system is $\underset{\_}{-3.40\text{eV}}$.

Explanation of Solution

Write the expression for the total energy.

    $E=K+U$                                                                                             (IX)

Conclusion:

Substitute $3.40\text{eV}$ for $K$ and $-6.80\text{eV}$ for $U$ in equation (IX) to find $E$.

    $\begin{array}{c}E=3.40\text{eV}+\left(-6.80\text{eV}\right)\\ =-3.40\text{eV}\end{array}$

Therefore, the total energy of the system is $\underset{\_}{-3.40\text{eV}}$.