Chapter 41, Problem 10P

(a)

To determine

Prove that the period of electron is $T=n^3{t}_0$ and calculate the numerical value of $t_0$.

Answer to Problem 10P

It is proved that $T=n^3{t}_0$ and the numerical value of $t_0$ is $152\text{ as}$.

Explanation of Solution

Write the equation for the angular momentum of electron.

    $L=m_{e}vr$

Here, $L$ is the angular momentum of electron, $m_e$ is the mass of electron, $v$ is the linear speed of electron, and $r$ is the radius of orbit.

Write the quantization condition for $L$.

    $L=n\hslash$

Here, $n$ is the principal quantum number, and $\hslash$ is the reduced Plank’s s constant.

Equate the right hand sides of above two equations.

    $m_{e}vr=n\hslash$

Rewrite the above expression in terms of $v$.

    $v=\frac{n\hslash }{m_{e}r}$

Write the equation for time period of orbital motion.

    $T=\frac{2\pi r}{v}$

Here, $T$ is the time period of orbital motion.

Rewrite the above expression by substituting $\frac{2\pi r}{v}$ for $T$.

    $\begin{array}{c}T=\frac{2\pi r}{\left(\frac{n\hslash }{m_{e}r}\right)}\\ =\frac{2\pi r{m}_{e}r}{n\hslash }\end{array}$

Write the equation to find the orbital radius.

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

Here, $r$ is the orbital radius, $k_e$ is the Coulomb’s constant, and $e$ is the magnitude of charge of electron.

Rewrite the expression for $T$ by substituting the above relation.

    $\begin{array}{c}T=\frac{2\pi r{m}_e\left(\frac{n^2{\hslash }^2}{m_e{k}_e{e}^2}\right)}{n\hslash }\\ =\frac{2\pi {\hslash }^3}{m_e{k}_e^2{e}^4}{n}^3\end{array}$

From dimensional analysis, it can be seen that the whole term $\frac{2\pi {\hslash }^3}{m_e{k}_e^2{e}^4}$ has the dimension of time. So it can be treated as $t_0$.

  $\begin{array}{l}T=\frac{2\pi {\hslash }^3}{m_e{k}_e^2{e}^4}{n}^3\\ T=t_0{n}^3\end{array}$                                                                                                           (I)

Thus, it is it is proved that $T=n^3{t}_0$.

Conclusion:

Substitute $3.14$ for $\pi$, $1.0546\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$, $\text{9}\text{.11}\times {10}^{-31}\text{kg}$ for $m_e$, $\text{8}\text{.99}\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $\text{1}\text{.602}\times {10}^{-19}\text{C}$ for $e$, and $1$ for $n$ in equation (I) to find $T$.

    $\begin{array}{c}T=\frac{2\left(3.14\right){(1.0546\times {10}^{-34}\text{ J}\cdot \text{s)}}^{\text{3}}}{\left(\text{9}\text{.11}\times {10}^{-31}\text{ kg}\right){\text{(8}\text{.99}\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\text{)}}^{\text{2}}{\text{(1}\text{.602}\times {10}^{-19}\text{ C)}}^{\text{4}}}{\left(1\right)}^3\\ =\frac{7.36\times {10}^{-102}}{4849.4\times {10}^{-89}}\text{s}\\ =1.52\times {10}^{-16}\text{ s}\left(\frac{1\text{as}}{{10}^{-18}\text{ s}}\right)\\ =152\text{as}\end{array}$

Therefore, it is proved that $T=n^3{t}_0$ and the numerical value of $t_0$ is $152\text{ as}$.

(b)

To determine

Number of revolutions made by electron in the excited state.

Answer to Problem 10P

Number of revolutions made by electron in the excited state is $8.23\times {10}^9$.

Explanation of Solution

Write the equation to find the number of revolutions made by electron in the excited state.

    $N=\frac{t_{life}}{T}$

Here, $N$ is the number of revolutions made by electron in the excited state and the $t_{life}$ is the lifetime in excited state.

Conclusion:

Substitute $3.14$ for $\pi$, $1.0546\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$, $\text{9}\text{.11}\times {10}^{-31}\text{kg}$ for $m_e$, $\text{8}\text{.99}\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $\text{1}\text{.602}\times {10}^{-19}\text{C}$ for $e$, and $2$ for $n$ in equation (I) to find $T$ for excited state.

  $\begin{array}{c}T=\frac{2\left(3.14\right){(1.0546\times {10}^{-34}\text{ J}\cdot \text{s)}}^{\text{3}}}{\left(\text{9}\text{.11}\times {10}^{-31}\text{ kg}\right){\text{(8}\text{.99}\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\text{)}}^{\text{2}}{\text{(1}\text{.602}\times {10}^{-19}\text{ C)}}^{\text{4}}}{\left(2\right)}^3\\ =\frac{7.36\times {10}^{-102}}{4849.4\times {10}^{-89}}\left(8\right)\text{s}\\ =1.22\times {10}^{-15}\text{s}\end{array}$

Substitute $10\text{μs}$ for $t_{life}$ and $1.22\times {10}^{-15}\text{s}$ for $T$ in the equation for $N$.

  $\begin{array}{c}N=\frac{10\text{μs}\left(\frac{{10}^{-6}\text{s}}{1\text{μs}}\right)}{1.22\times {10}^{-15}\text{s}}\\ =8.23\times {10}^9\end{array}$

Therefore, the number of revolutions made by electron in the excited state is $8.23\times {10}^9$.

(c)

To determine

Check whether is it possible to treat the electron in $n=2$ as “living for a long time” on considering the period of revolution of electron as a year.

Answer to Problem 10P

On comparing the period of revolution electron with that of Sun, it is correct that electron in $n=2$ is “living for a long time”.

Explanation of Solution

It is found that in part (b), $T$ for electron at $n=2$ is $1.22\times {10}^{-15}\text{s}$. Electron makes $8.23\times {10}^9$ revolutions in $16\text{μs}$. If it is assumed that the period of electron as one year, electron stays at $n=2$ for $8.23\times {10}^9$ years. It is a very big value compared to the period of revolution of Sun.

Conclusion:

Therefore, on comparing the period of revolution electron with that of Sun, it is correct that electron in $n=2$ is “living for a long time”.