Bohr's Model — How Electrons Orbit

The Problem Bohr Was Solving

By 1913, physicists were in trouble. Rutherford’s nuclear model (1911) had electrons orbiting the nucleus like planets orbiting the sun. But classical electromagnetic theory said this couldn’t work — an accelerating charged particle radiates energy, so the electron should spiral into the nucleus in about $10^{-10}$ seconds. Clearly, atoms don’t collapse like that.

Simultaneously, experimentalists had recorded the hydrogen spectrum: hydrogen emits light only at specific wavelengths — a series of sharp lines, not a continuous rainbow. Classical physics couldn’t explain why.

Niels Bohr solved both problems with one bold model in 1913.

Bohr’s Four Postulates

Postulate 1 — Stationary orbits: Electrons orbit the nucleus in specific circular paths called stationary orbits or allowed orbits. While in these orbits, the electron does NOT emit radiation (defying classical electromagnetism).

Postulate 2 — Quantization of angular momentum: Only those orbits are allowed for which the electron’s angular momentum is an integer multiple of $\frac{h}{2\pi}$ :

mvr = \frac{nh}{2\pi} = n\hbar \quad (n = 1, 2, 3, \ldots)

where $n$ is the principal quantum number and $h$ is Planck’s constant.

Postulate 3 — Energy transition: An electron can jump between orbits by absorbing or emitting a photon. If it jumps from higher orbit ( $E_i$ ) to lower ( $E_f$ ):

h\nu = E_i - E_f

Postulate 4 — Stability: The innermost orbit ( $n=1$ , ground state) is the most stable. Electrons remain there unless disturbed.

Bohr’s key genius was Postulate 1 — simply asserting that electrons in allowed orbits don’t radiate, against classical physics. He couldn’t explain WHY, but it worked. The WHY came later from quantum mechanics (de Broglie waves forming standing waves in orbits).

Deriving the Bohr Radius

For the electron to stay in a circular orbit, the Coulomb attraction must provide the centripetal force:

\frac{ke^2}{r^2} = \frac{mv^2}{r}

Combined with the angular momentum quantization:

mvr = \frac{nh}{2\pi}

Solving these two equations simultaneously gives:

r_n = \frac{n^2 a_0}{Z}

a_0 = 0.529 \text{ Å} = 0.529 \times 10^{-10} \text{ m} \quad (\text{Bohr radius})

$n$ = principal quantum number
$Z$ = atomic number (charge on nucleus)
For hydrogen, $Z = 1$ : $r_1 = 0.529$ Å, $r_2 = 4 \times 0.529 = 2.116$ Å

Radii scale as $n^2$ : the second orbit is 4 times farther than the first, the third is 9 times.

Energy of Bohr Orbits

The total energy (kinetic + potential) of the electron in orbit $n$ :

E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV}

For hydrogen ( $Z = 1$ ):

$E_1 = -13.6$ eV (ground state)
$E_2 = -3.4$ eV
$E_3 = -1.51$ eV
$E_\infty = 0$ eV (electron completely removed)

Why negative? The zero of energy is defined as the electron at rest at infinity. Bound electrons have negative total energy because they’ve been captured and must have energy added to escape.

Ionization energy of hydrogen = 13.6 eV (energy needed to remove the electron from ground state to infinity).

JEE Main almost every year asks: “What is the energy needed to excite hydrogen from $n=1$ to $n=3$ ?” The answer: $E_3 - E_1 = -1.51 - (-13.6) = 12.09$ eV. Note the sign: excitation requires energy input (photon absorbed).

The Hydrogen Spectrum

When an electron falls from orbit $n_2$ to orbit $n_1$ (where $n_2 > n_1$ ), it emits a photon:

\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

$R_H = 1.097 \times 10^7$ m $^{-1}$ (Rydberg constant)

Spectral Series of Hydrogen

Series	Transition to $n_1$	Region	Key lines
Lyman	$n_1 = 1$	UV	First line: $n=2 \to 1$
Balmer	$n_1 = 2$	Visible	H $\alpha$ (red), H $\beta$ (blue-green)
Paschen	$n_1 = 3$	IR	—
Brackett	$n_1 = 4$	Far IR	—
Pfund	$n_1 = 5$	Far IR	—

Memory trick for series order: Lyman, Balmer, Paschen, Brackett, Pfund → “Live Boring Physics But Pass.” Lyman is in UV (lowest), Balmer is visible, rest are in IR (higher energy = shorter wavelength, remember $E = hc/\lambda$ ).

Number of Spectral Lines

If an electron falls from orbit $n$ to ground state (passing through intermediate states), the maximum number of spectral lines emitted:

\text{Lines} = \frac{n(n-1)}{2}

From $n=4$ : maximum $\frac{4 \times 3}{2} = 6$ lines (transitions 4→3, 4→2, 4→1, 3→2, 3→1, 2→1).

Velocity of Electron in Bohr’s Orbits

From the angular momentum and orbital mechanics:

v_n = \frac{Ze^2}{2\varepsilon_0 nh} = \frac{\alpha c Z}{n}

where $\alpha = \frac{1}{137}$ is the fine structure constant and $c$ is the speed of light.

For hydrogen ground state: $v_1 = \frac{c}{137} \approx 2.18 \times 10^6$ m/s. This is about 0.7% of the speed of light — justifying the non-relativistic treatment.

Velocity decreases as $\frac{1}{n}$ : outer orbits are slower.

Limitations of Bohr’s Model

Bohr’s model was a major step forward but is incomplete:

Works only for hydrogen-like ions ( $H$ , $He^+$ , $Li^{2+}$ , etc.) — fails for multi-electron atoms
Cannot explain fine structure — spectral lines are actually closely spaced doublets or multiplets
Cannot explain Zeeman effect — splitting of lines in magnetic fields
Cannot explain the intensity of spectral lines (why some transitions are more probable)
Electrons don’t actually move in circular orbits — quantum mechanics replaced this with probability clouds (orbitals)

Solved Examples

Example 1 — CBSE Level

Find the radius of the second Bohr orbit in hydrogen.

r_2 = \frac{n^2 a_0}{Z} = \frac{4 \times 0.529}{1} = 2.116 \text{ Å}

Example 2 — JEE Main Level

What is the wavelength of the first line of the Balmer series?

First Balmer line: $n_2 = 3 \to n_1 = 2$

\frac{1}{\lambda} = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 \times 10^7 \times \frac{5}{36}

\frac{1}{\lambda} = 1.524 \times 10^6 \text{ m}^{-1}

$\lambda = 656 \text{ nm} \quad \text{(red light --- H}\alpha \text{ line)}$

Example 3 — JEE Advanced Level

A hydrogen atom in state $n=4$ de-excites to ground state. How many different photons can be emitted? What are the series of each transition?

Transitions: $4\to 3$ (Paschen), $4\to 2$ (Balmer), $4\to 1$ (Lyman), $3\to 2$ (Balmer), $3\to 1$ (Lyman), $2\to 1$ (Lyman). Total: 6 photons.

Lyman: 3 transitions ( $4\to 1$ , $3\to 1$ , $2\to 1$ ) Balmer: 2 transitions ( $4\to 2$ , $3\to 2$ ) Paschen: 1 transition ( $4\to 3$ )

Common Mistakes to Avoid

Mistake 1: Writing $E_n = +\frac{13.6}{n^2}$ eV (positive). Energy of bound states is always negative. The ionization energy is the amount you need to ADD to reach zero energy.

Mistake 2: Confusing absorption and emission. Electron going to higher $n$ = absorbs photon. Electron going to lower $n$ = emits photon. The frequency/wavelength formula $\Delta E = h\nu$ is the same for both.

Mistake 3: Using $r_n = n^2 a_0$ for all atoms. For hydrogen-like ions, $r_n = \frac{n^2 a_0}{Z}$ . In helium ( $Z=2$ ), the ground state radius is $\frac{0.529}{2} = 0.265$ Å.

Mistake 4: Forgetting that the Lyman series is in the UV range, not visible. Students sometimes confuse Lyman and Balmer. Balmer is visible (the one that’s photographically observed easily).

Practice Questions

Q1. Calculate the energy emitted when hydrogen goes from $n=3$ to $n=1$ .

$E_3 - E_1 = -1.51 - (-13.6) = 12.09$ eV emitted (transition from higher to lower energy).

Q2. An electron in $He^+$ is in the second orbit. Find the orbital radius.

$r_2 = \frac{n^2 a_0}{Z} = \frac{4 \times 0.529}{2} = 1.058$ Å

Q3. Find the frequency of the photon that ionizes hydrogen from the ground state.

Energy needed = 13.6 eV $= 13.6 \times 1.6 \times 10^{-19} = 2.176 \times 10^{-18}$ J. $\nu = \frac{E}{h} = \frac{2.176 \times 10^{-18}}{6.626 \times 10^{-34}} = 3.28 \times 10^{15}$ Hz.

Q4. A photon of 12.09 eV is absorbed by a hydrogen atom in the ground state. To which orbit does the electron go?

Energy of orbit: $E_n = E_1 + 12.09 = -13.6 + 12.09 = -1.51$ eV. This matches $E_3 = -\frac{13.6}{9} = -1.51$ eV. The electron goes to $n = 3$ .

Additional Worked Examples

Example 4 — Energy Level Transitions

For a hydrogen-like ion $Li^{2+}$ ( $Z=3$ ), find the energy of the electron in the third orbit and the wavelength of light emitted when it transitions from $n=3$ to $n=1$ .

Using $E_n = -\frac{13.6 \, Z^2}{n^2}$ eV with $Z = 3$ , $n = 3$ :

E_3 = -\frac{13.6 \times 9}{9} = -13.6 \text{ eV}

E_1 = -\frac{13.6 \times 9}{1} = -122.4 \text{ eV}

\Delta E = E_3 - E_1 = -13.6 - (-122.4) = 108.8 \text{ eV}

Using $E = hc/\lambda$ and $hc = 1240$ eV·nm:

\lambda = \frac{1240}{108.8} = 11.4 \text{ nm}

This falls in the extreme UV range — much shorter wavelength than hydrogen because $Z=3$ means tighter binding.

Example 5 — Speed and Time Period

Find the time period of revolution of an electron in the $n=2$ orbit of hydrogen.

v_n = \frac{c}{137 \cdot n} = \frac{3 \times 10^8}{137 \times 2} = 1.09 \times 10^6 \text{ m/s}

r_2 = 4 \times 0.529 \times 10^{-10} = 2.116 \times 10^{-10} \text{ m}

T = \frac{2\pi r_2}{v_2} = \frac{2\pi \times 2.116 \times 10^{-10}}{1.09 \times 10^6} = 1.22 \times 10^{-15} \text{ s}

The electron completes about $10^{15}$ orbits per second — incredibly fast.

T_n = \frac{2\pi r_n}{v_n} \propto \frac{n^3}{Z^2}

f_n = \frac{1}{T_n} \propto \frac{Z^2}{n^3}

Time period increases as $n^3$ — outer orbits are much slower in angular speed.

NEET 2023 asked: “The ratio of speeds of electrons in the first and second Bohr orbits of hydrogen is?” Since $v_n \propto 1/n$ , the ratio is $v_1 : v_2 = 2 : 1$ . Similarly, the ratio of radii is $r_1 : r_2 = 1 : 4$ (since $r_n \propto n^2$ ). These proportionality relations are tested frequently.

Q5. For hydrogen, calculate the ratio of the kinetic energy to the potential energy in any Bohr orbit.

In the $n$ -th orbit, kinetic energy $KE = -E_n = \frac{13.6}{n^2}$ eV, and potential energy $PE = 2E_n = -\frac{27.2}{n^2}$ eV.

Ratio $\frac{KE}{PE} = \frac{13.6/n^2}{-27.2/n^2} = -\frac{1}{2}$

The kinetic energy is always half the magnitude of the potential energy, with opposite sign. This is the virial theorem applied to the Coulomb force.

Q6. How many spectral lines are emitted when an electron in hydrogen falls from $n=5$ to the ground state?

Maximum spectral lines $= \frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10$ lines. These include: 3 Lyman lines ( $5 \to 1$ , $4 \to 1$ , $3 \to 1$ , $2 \to 1$ — actually 4 Lyman lines), 3 Balmer lines ( $5 \to 2$ , $4 \to 2$ , $3 \to 2$ ), 2 Paschen lines ( $5 \to 3$ , $4 \to 3$ ), and 1 Brackett line ( $5 \to 4$ ). Total = 4 + 3 + 2 + 1 = 10. Correct.

FAQs

Q: If electrons don’t really move in circular orbits, what DO they do?

Quantum mechanics describes electrons as probability clouds called orbitals. The electron has a probability of being at various locations around the nucleus. Bohr’s circular orbits are a simplified model that gives correct energy levels for hydrogen but incorrect physical picture.

Q: Why is the ground state energy of hydrogen $-13.6$ eV and not some other value?

It’s determined by the fundamental constants: $E_1 = -\frac{me^4}{8\varepsilon_0^2 h^2}$ . Plugging in the values gives exactly $-13.6$ eV. This agreement between theory and experiment was one of Bohr’s greatest triumphs.

Q: What is the relation between Bohr orbits and the periodic table?

The principal quantum number $n$ corresponds roughly to the “shell” of electrons. $n=1$ is K-shell (max 2 electrons), $n=2$ is L-shell (max 8), $n=3$ is M-shell (max 18). This is why elements in the same period have the same outermost $n$ value.

Q: Can Bohr’s model be applied to molecules?

No. Bohr’s model only applies to hydrogen-like species (one electron orbiting one nucleus). For multi-electron atoms, electron-electron repulsions change everything. For molecules, we need molecular orbital theory — far beyond Bohr’s model.