Atomic Models

Lessons

Senior 5 · Physics

Atomic Models

4 lessons

Lesson 2 of 4

Bohr's Model of the Hydrogen Atom

25 min

A rule that should not work

By 1913 physicists knew two things that did not fit together. Rutherford's nucleus was certain, but an orbiting electron should radiate its energy away and crash into it in a hundred-billionth of a second. And Max Planck (1900) and Albert Einstein (1905) had shown that light energy comes in packets, photons, of energy $E = h f$ . The Danish physicist Niels Bohr made a bold move: he simply declared that the electron in a hydrogen atom can occupy only certain orbits, in which it does not radiate. The rule had no deeper justification at the time, but from it Bohr calculated the size of the hydrogen atom and the exact colours of the light it gives out. Those predictions matched experiment to better than one part in a thousand.

Bohr's postulates

The electron moves round the nucleus in certain circular orbits only, called stationary states. While in one of them it does not radiate energy.
In an allowed orbit, the angular momentum of the electron is a whole-number multiple of $\frac{h}{2\pi}$ : $mvr = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \ldots$ where $h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$ is the Planck constant and $n$ is the principal quantum number.
The atom emits or absorbs radiation only when the electron jumps from one allowed orbit to another. The energy of the single photon emitted or absorbed equals the difference in energy of the two states: $hf = E_{\text{upper}} - E_{\text{lower}}$

Radius, speed and energy of the orbits

For a hydrogen atom the nucleus is a single proton of charge $+ e$ . The electric attraction provides the centripetal force:

\frac{e^2}{4\pi\varepsilon_0r^2} = \frac{mv^2}{r} \quad (1)

From postulate 2, $v = \frac{nh}{2\pi mr}$ . Substituting into (1) and solving for $r$ :

r_n = \frac{\varepsilon_0h^2}{\pi me^2}\,n^2 = n^2a_0, \qquad a_0 = \frac{\varepsilon_0h^2}{\pi me^2} = 5.29 \times 10^{-11}\,\mathrm{m}

$a_{0}$ , the radius of the smallest orbit, is the Bohr radius: it gives a hydrogen atom a diameter of about 0.1 nm, as measured. The speed in the $n$ th orbit is $v_n = \frac{e^2}{2\varepsilon_0nh} = \frac{2.19 \times 10^6}{n}\,\mathrm{m\,s^{-1}}$ .

The kinetic energy is $\frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0r}$ (from (1)), and the electric potential energy is $-\frac{e^2}{4\pi\varepsilon_0r}$ , so the total energy is

E_n = -\frac{e^2}{8\pi\varepsilon_0r_n} = -\frac{me^4}{8\varepsilon_0^2h^2}\cdot\frac{1}{n^2} = -\frac{13.6\,\mathrm{eV}}{n^2}

( $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$ is the energy an electron gains when accelerated through 1 volt.) This has exactly the same form as a satellite's energy in Unit 5: negative, because the electron is bound, and equal to minus its kinetic energy.

Definition

An energy level is one of the allowed total energies of an electron in an atom. For hydrogen, $E_n = -\frac{13.6}{n^2}\,\mathrm{eV}$ . The lowest level ( $n = 1$ ) is the ground state; higher levels are excited states. The ionisation energy is the energy needed to remove the electron completely from the ground state: 13.6 eV for hydrogen. Example: $E_2 = -\frac{13.6}{4} = -3.40\,\mathrm{eV}$ .

The energy level diagram

An energy level diagram draws each allowed energy as a horizontal line on a vertical energy scale. Zero is at the top ( $n = \infty$ ), where the electron is just free of the atom. The levels crowd together as $n$ increases.

To lift an electron from a lower to a higher level, the atom must absorb exactly the right amount of energy, from a photon of exactly that energy or from a collision with another particle (for example an electron in a discharge tube). An excited atom soon falls back, giving out photons as the electron drops down, in one or several steps.

Worked example 1: the second orbit

For the $n = 2$ state of hydrogen, find (a) the orbit radius, (b) the electron's speed, (c) its total energy, (d) its kinetic and potential energies.

Solution

(a) $r_2 = 2^2 \times 5.29 \times 10^{-11} = 2.12 \times 10^{-10}\,\mathrm{m}$ .

(b) $v_2 = \frac{2.19 \times 10^6}{2} = 1.09 \times 10^6\,\mathrm{m\,s^{-1}}$ , well below the speed of light.

(d) $E_k = -E_2 = +3.40\,\mathrm{eV}$ ; $E_p = 2E_2 = -6.80\,\mathrm{eV}$ . Check: $3.40 + (-6.80) = -3.40\,\mathrm{eV}$ .

Worked example 2: exciting and ionising

(a) What energy is needed to excite a hydrogen atom from its ground state to $n = 3$ ? (b) An electron with 12.5 eV of kinetic energy collides with a hydrogen atom in its ground state. Which levels can the atom reach, and how much energy does the electron keep in each case? (c) What is the least energy needed to ionise a hydrogen atom that is already in $n = 2$ ?

Solution

(a) $E_3 - E_1 = -1.51 - (-13.6) = 12.1\,\mathrm{eV}$ .

(b) $n = 2$ needs 10.2 eV and $n = 3$ needs 12.1 eV, both below 12.5 eV; $n = 4$ needs 12.75 eV, too much. A colliding electron can give part of its energy, so the atom can go to $n = 2$ (the electron keeps 2.3 eV) or $n = 3$ (it keeps 0.4 eV). A photon is different: it is absorbed only if its energy matches a gap exactly.

Successes and limits of Bohr's model

Successes	Limitations
explains why atoms are stable and have a definite size	works only for hydrogen and one-electron ions such as He⁺; fails for atoms with two or more electrons
predicts the wavelengths of the hydrogen spectrum exactly (Lesson 3)	cannot explain why some spectral lines are brighter than others, or why many lines are split into close pairs
gives the correct ionisation energy, 13.6 eV	gives no reason why angular momentum is quantised; electrons do not really follow sharp circular paths

In the 1920s the Bohr model was replaced by quantum mechanics, in which the electron is described by a wave and the "orbits" become regions of probability. Remarkably, quantum mechanics gives exactly the same energy levels for hydrogen, $-\frac{13.6}{n^2}$ eV, so the energy level diagram above remains correct.

Warning

Common mistakes. (1) Dropping the minus sign: energy levels are negative, and a higher level has a less negative energy. (2) Writing $r \propto n$ : the radius grows as $n^{2}$ . (3) Forgetting to convert eV to joules ( $\times 1.60 \times 10^{-19}$ ) before using $E = h f$ with $h$ in J s. (4) Thinking a photon can give part of its energy to an atom: a photon is absorbed whole or not at all, while a colliding electron can share its energy.

Tip

Learn three numbers for hydrogen and derive the rest: $E_1 = -13.6\,\mathrm{eV}$ , $r_1 = 0.053\,\mathrm{nm}$ , $v_1 = 2.2 \times 10^6\,\mathrm{m\,s^{-1}}$ . Then $E_n = \frac{E_1}{n^2}$ , $r_{n} = n^{2} r_{1}$ , $v_n = \frac{v_1}{n}$ .

Bohr model energy levels, derivation using physics (Khan Academy). Watch how the force equation and the quantised angular momentum combine to give energies proportional to −1 ÷ n².

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Practice

State Bohr's three postulates.
Starting from $\frac{e^2}{4\pi\varepsilon_0r^2} = \frac{mv^2}{r}$ and $mvr = \frac{nh}{2\pi}$ , show that $r \propto n^2$ .
Calculate the energies of the $n = 4$ and $n = 5$ levels of hydrogen.
Find the energy, in eV and in joules, needed to excite a hydrogen atom from $n = 2$ to $n = 4$ .
Find the radius of the $n = 5$ orbit and compare it with the ground-state radius.
Show that the kinetic energy of the electron in a Bohr orbit equals minus its total energy.
A hydrogen atom in its ground state is hit by a photon of energy 11.0 eV. Will the photon be absorbed? Explain.
Evaluate the statement: "Bohr's model has been replaced, so it is useless."

Show answersHide answers

Electrons move in certain allowed orbits without radiating; in these orbits $mvr = \frac{nh}{2\pi}$ ; radiation is emitted or absorbed only in a jump between orbits, with $hf = \Delta E$ .
$v = \frac{nh}{2\pi mr}$ into $\frac{e^2}{4\pi\varepsilon_0r} = mv^2$ gives $\frac{e^2}{4\pi\varepsilon_0r} = \frac{n^2h^2}{4\pi^2mr^2}$ , so $r = \frac{\varepsilon_0h^2n^2}{\pi me^2} \propto n^2$ .
$E_4 = -\frac{13.6}{16} = -0.85\,\mathrm{eV}$ ; $E_5 = -\frac{13.6}{25} = -0.54\,\mathrm{eV}$ .
$-0.85 - (-3.40) = 2.55\,\mathrm{eV} = 2.55 \times 1.60 \times 10^{-19} = 4.08 \times 10^{-19}\,\mathrm{J}$ .
$r_5 = 25 \times 0.0529 = 1.32\,\mathrm{nm}$ , 25 times the ground-state radius.
From the force equation, $\frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0r}$ ; the potential energy is $-\frac{e^2}{4\pi\varepsilon_0r}$ ; the total is $-\frac{e^2}{8\pi\varepsilon_0r} = -E_k$ .
No. The gaps from the ground state are 10.2 eV (to $n = 2$ ) and 12.1 eV (to $n = 3$ ); 11.0 eV matches no gap, and a photon cannot give only part of its energy, so it passes through.
Not true. Its ideas of stationary states, quantised energy levels and photons from jumps between them are still correct, and its energy levels for hydrogen agree with quantum mechanics. It fails only for many-electron atoms and finer details.

Key points

Bohr: stationary orbits, $mvr = \frac{nh}{2\pi}$ , and $hf = E_{\text{upper}} - E_{\text{lower}}$ .
Hydrogen: $r_n = n^2 \times 0.0529\,\mathrm{nm}$ , $v_n = \frac{2.19 \times 10^6}{n}\,\mathrm{m\,s^{-1}}$ , $E_n = -\frac{13.6}{n^2}\,\mathrm{eV}$ .
Ground state $n = 1$ ; ionisation energy 13.6 eV.
The model works for hydrogen-like atoms only; quantum mechanics replaced it but keeps the same energy levels.

Sources and credits

6.4 Bohr's Model of the Hydrogen Atom, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Bohr model energy levels (derivation using physics), Khan Academy, embedded from YouTube.

The two diagrams, the worked examples and the practice exercises are written by Ganzaa.

1A rule that should not work

2Bohr's postulates

3Radius, speed and energy of the orbits

4The energy level diagram

5Successes and limits of Bohr's model

A rule that should not work

Bohr's postulates

Radius, speed and energy of the orbits

The energy level diagram

Successes and limits of Bohr's model