Chapter 8, Problem 120FP

Interpretation Introduction

(a)

Interpretation:

An expression for the total energy, E of the electron (mass m_e) moving in a circular orbit of radius r with speed U should be written.

Concept introduction:

An electron orbiting in a circular orbit around the nucleus experiences two types of energies. They are kinetic energy and the potential energy due to attractive forces between nucleus and electron.

Interpretation Introduction

(b)

Interpretation:

It should be shown that the total energy of the electron is $E=-\frac{e^2}{8\pi {\epsilon }_{0}r}$ using the condition that the force of attraction between the electron and proton has the same magnitude as the centrifugal force $\frac{m_e{U}^2}{r}$.

Concept introduction:

Electrostatic force between two charges is given by the following relation:

$F=k_e\frac{q_1{q}_2}{r^2}$

Here, k_e − Coulomb constant

q₁ and q₂ − magnitude of the charges

r − distance between the two charges.

Interpretation Introduction

(c)

Interpretation:

It should be shown that the energy and radius of the nth orbit are respectively, $E_n=-\frac{R_{\infty }}{n^2}$ and

$r_n=a_0\times n^2$ with $R_{\infty }=\frac{m_e{e}^4}{8{\epsilon }_0^2{h}^2}=2.17987\times {10}^{-18}\text{ J}$ and $a_0=\frac{h^2{\epsilon }_0}{\pi m_e{e}^2}=5.29177\times {10}^{11}\text{ m}$

Concept introduction:

Using the condition that the force of attraction between the electron and proton has the same magnitude as the centrifugal force:

$k_e\frac{q_1{q}_2}{r^2}=\frac{m_e{U}^2}{r}$

The total energy of the electron is as follows:

$E=-\frac{e^2}{8\pi {\epsilon }_{0}r}$

Interpretation Introduction

(d)

Interpretation:

$R_{\infty }$ should be converted to R_H by replacing m_e in the expression for $R_{\infty }$ with so called reduced mass µ.

Concept introduction:

Reduced mass can be calculated as follows:

$\mu =\frac{m_e{m}_p}{\left(m_e+m_p\right)}$

Here, $m_p$ is mass of proton and $m_e$ is mass of electron.

Also,

$\begin{array}{l}{m}_p=1.67262\times {10}^{-27}\text{ kg}\\ m_e=9.10938\times {10}^{-31}\text{ kg}\end{array}$

The conversion of $R_{\infty }$ to R_H corrects for the fact that, because the proton is not significantly large compared to the electron, the nucleus is not actually stationary.