Chapter 29, Problem 10P

(a)

To determine

The uncertainty in momentum of electron in terms of $r$.

Answer to Problem 10P

The uncertainty in momentum is $\Delta p \ge \frac{\hslash }{2r}$.

Explanation of Solution

Write the equation for uncertainty principle.

    $\Delta x\Delta p \ge \frac{\hslash }{2}$

Here, $\Delta x$ is the uncertainty in position, $\text{}\Delta \text{}p$ is the uncertainty in momentum and $\hslash$ is the reduced Plank’s constant.

Conclusion:

Rewrite the above relation by substituting $r$ for $x$.

    $\left(r\right)\Delta p \ge \frac{\hslash }{2}$

Rewrite the above relation in terms of $\text{}\Delta \text{}p$.

    $\Delta p \ge \frac{\hslash }{2\left(r\right)}$

Therefore, the uncertainty in momentum is $\Delta p \ge \frac{\hslash }{2r}$.

(b)

To determine

The kinetic energy of electron.

Answer to Problem 10P

The kinetic energy of electron will be $\frac{{\hslash }^2}{2m_e{r}^2}$.

Explanation of Solution

Write the equation for momentum.

    $K=\frac{{\left(\Delta p\right)}^2}{2m_e}$

Here, $K$ is the kinetic energy and $m_e$ is the mass of electron.

Conclusion:

Rewrite the above equation by substituting $\frac{\hslash }{2r}$ for $\Delta p$.

  $\begin{array}{c}K=\frac{{\left( \frac{\hslash }{2r}\right)}^2}{2m_e}\\ =\frac{{\hslash }^2}{2m_e{r}^2}\end{array}$

Therefore, the kinetic energy of electron will be $\frac{{\hslash }^2}{2m_e{r}^2}$.

(c)

To determine

The total energy of electron in terms of $r$.

Answer to Problem 10P

The total energy of electron in terms of $r$ is $\frac{{\hslash }^2}{2m_e{r}^2}-\frac{k_e{e}^2}{r}$.

Explanation of Solution

Write the equation for total energy of electron.

    $E=K+U$

Here, $E$ is the total energy and $U$ is the potential energy.

Write the equation for $U$.

  $U=-\frac{k_e{e}^2}{r}$

Here, $k$ is the electrostatic constant and $e$ is the magnitude of electric charge.

Rewrite the equation for $E$ by substituting the above relation for $U$ and $\frac{{\hslash }^2}{2m_e{r}^2}$ for $K$.

Conclusion:

Rewrite the above equation by substituting $\frac{\hslash }{2r}$ for $\Delta p$.

  $\begin{array}{c}E=\frac{{\hslash }^2}{2m_e{r}^2}+\left(-\frac{k_e{e}^2}{r}\right)\\ =\frac{{\hslash }^2}{2m_e{r}^2}-\frac{k_e{e}^2}{r}\end{array}$                                                                                   (I)

Therefore, the total energy of electron in terms of $r$ is $\frac{{\hslash }^2}{2m_e{r}^2}-\frac{k_e{e}^2}{r}$.

(d)

To determine

The value of $r$.

Answer to Problem 10P

The value of $r$ is $\frac{{\hslash }^2}{m_e{k}_e{e}^2}$.

Explanation of Solution

Write the condition for the value of $E$ to be minimum.

    $\frac{dE}{dr}=0$

Conclusion:

Rewrite the above equation by substituting for $\frac{{\hslash }^2}{2m_e{r}^2}-\frac{k_e{e}^2}{r}$ for $E$.

  $\begin{array}{c}\frac{dE}{dr}=0\\ \frac{d\left(\frac{{\hslash }^2}{2m_e{r}^2}-\frac{k_e{e}^2}{r}\right)}{dr}=0\\ -\frac{{\hslash }^2}{m_e{r}^3}+\frac{k_e{e}^2}{r^2}=0\end{array}$

Rewrite the above relation in terms of $r$.

  $r=\frac{{\hslash }^2}{m_e{k}_e{e}^2}$

The calculated value $\frac{{\hslash }^2}{m_e{k}_e{e}^2}$ is the Bohr radius.

Therefore, the value of $r$ is $\frac{{\hslash }^2}{m_e{k}_e{e}^2}$.

(e)

To determine

The resulting total energy.

Answer to Problem 10P

The resulting total energy is $-13.6\text{eV}$.

Explanation of Solution

Write the equation for $E$ found in part (c).

    $E=\frac{{\hslash }^2}{2m_e{r}^2}-\frac{k_e{e}^2}{r}$

Rewrite the above equation by substituting $\frac{{\hslash }^2}{m_e{k}_e{e}^2}$ for $r$.

    $\begin{array}{c}E=\frac{{\hslash }^2}{2m_e{\left(\frac{{\hslash }^2}{m_e{k}_e{e}^2}\right)}^2}-\frac{k_e{e}^2}{\left(\frac{{\hslash }^2}{m_e{k}_e{e}^2}\right)}\\ =-\frac{m_e{k}_e^2{e}^4}{2{\hslash }^2}\end{array}$

Conclusion:

Substitute $9.109\times {10}^{-31}\text{kg}$ for $m_e$, $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k_e$, $1.602\times {10}^{-19}\text{ C}$ for $e$, and $1.055\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ in the above equation to find $E$.

    $\begin{array}{c}E=-\frac{\left(9.109\times {10}^{-31}\text{kg}\right){\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^4}{2{\left(1.055\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}\\ =-2.179\times {10}^{-18}\text{ J}\left(\frac{1\text{ eV}}{1.602\times {10}^{-19}\text{ J}}\right)\\ =-13.6\text{eV}\end{array}$

Therefore, the resulting total energy is $-13.6\text{eV}$.

(f)

To determine

States that the answers are in agreement the conclusions of Bohr theory.

Explanation of Solution

The uncertainty of momentum of electron calculated in part (a) is $\Delta p \ge \frac{\hslash }{2r}$. It is two times the minimum possible value of momentum. This makes the result is in agreement with the conclusions of Bohr theory.