Chemistry · Structure of Atom

Bohr's Model for Hydrogen Atom

In 1913 Niels Bohr was the first to explain quantitatively the general features of the hydrogen atom and its spectrum, blending Rutherford's nucleus with Planck's quantisation of energy. NCERT Class 11 Chemistry §2.4 builds the model from four postulates and derives the orbit radius, the energy levels and the line spectrum. For NEET this subtopic is high-yield: radius, energy and spectral-series calculations appear almost every year, including a fresh wavelength-ratio problem in 2025.

Why Bohr Reworked the Atom

Rutherford's nuclear model placed the electron in orbit around a tiny dense nucleus, but classical electromagnetism predicted that an orbiting (accelerating) charge must continuously radiate energy and spiral into the nucleus within a fraction of a second. That model could not explain why atoms are stable, nor why excited hydrogen emits light only at certain discrete frequencies rather than a continuous band.

When an electric discharge is passed through gaseous hydrogen, the $\ce{H2}$ molecules dissociate and the energetically excited $\ce{H}$ atoms emit electromagnetic radiation of discrete frequencies. Passed through a prism, this light separates into a set of sharp lines — the hydrogen line spectrum — rather than a rainbow. Bohr's central achievement was to make a model that produces exactly such a discrete spectrum.

NEET Trap

Line spectrum, not continuous spectrum

A glowing solid (a bulb filament) gives a continuous spectrum; an excited gaseous atom gives a line spectrum. Examiners exploit this: hydrogen produces a line emission spectrum precisely because its energy levels are quantised.

Discrete energy levels → discrete photon energies → sharp lines.

The Four Postulates

Bohr's model for the hydrogen atom rests on four postulates. The first two introduce stationary states; the third connects energy gaps to radiation; the fourth quantises angular momentum, which is the engine that produces all the quantitative results.

PostulateStatement
1. Stationary orbitsThe electron moves around the nucleus in circular paths of fixed radius and energy, called orbits, stationary states or allowed energy states, arranged concentrically about the nucleus.
2. Constant energyThe energy of an electron in an orbit does not change with time. The electron jumps to a higher state only when it absorbs the required energy, and emits energy when it falls to a lower state. The change is not continuous.
3. Bohr frequency ruleWhen a transition occurs between two states differing in energy by $\Delta E$, the frequency of the radiation absorbed or emitted is $\nu = \dfrac{E_2 - E_1}{h}$.
4. Quantised angular momentumThe angular momentum of the electron is an integral multiple of $h/2\pi$: $\; m_e v r = n\dfrac{h}{2\pi}$, with $n = 1, 2, 3, \dots$

The angular momentum used in postulate 4 follows from treating the electron of mass $m_e$ on a circular path of radius $r$ as a rotating body. With moment of inertia $I = m_e r^2$ and angular velocity $\omega = v/r$,

$$\text{angular momentum} = I \times \omega = m_e r^2 \times \frac{v}{r} = m_e v r.$$

Quantising this quantity — allowing it to take only the values $n h / 2\pi$ — is the single radical assumption. Radiation is emitted or absorbed only when the electron jumps from one quantised value of angular momentum to another, so Maxwell's classical electromagnetic theory does not apply and only certain fixed orbits are allowed. This idea grew directly out of Planck's quantisation of energy; for the experimental clues that pushed physics toward it, see developments leading to Bohr's model.

+ n = 1 n = 1 n = 2 n = 3 r₂ = 4r₁ r₃ = 9r₁ absorption (jump up)
Figure 1. Bohr's concentric circular orbits. The electron occupies a stationary state of fixed radius and energy and absorbs a photon to jump outward (or emits one to fall inward). Radii scale as $n^2$, so the second orbit is four times and the third nine times the first.

Radius of the Stationary States

Bohr's theory gives the radii of the allowed orbits of hydrogen as

$$r_n = n^2 a_0, \qquad a_0 = 52.9\ \text{pm}.$$

The first stationary state ($n = 1$), called the Bohr orbit, has radius 52.9 pm, and the hydrogen electron is normally found here. As $n$ increases the radius increases as $n^2$, so the electron sits progressively farther from the nucleus. The stationary states are numbered $n = 1, 2, 3, \dots$; these integers are the principal quantum numbers.

Orbit nFactor n²Radius (H, Z = 1)
1152.9 pm
24211.6 pm
39476.1 pm
416846.4 pm

Energy of the Stationary States

The most important property of a stationary state is its energy. For hydrogen,

$$E_n = -R_H\left(\frac{1}{n^2}\right), \qquad R_H = 2.18 \times 10^{-18}\ \text{J}.$$

Here $R_H$ is the Rydberg constant expressed as an energy. The ground state ($n = 1$) is the lowest and most stable: $E_1 = -2.18 \times 10^{-18}\,(1/1^2) = -2.18 \times 10^{-18}$ J. For $n = 2$, $E_2 = -2.18 \times 10^{-18}\,(1/2^2) = -0.545 \times 10^{-18}$ J. In electron-volts the same expression reads $E_n = -13.6/n^2$ eV, so $E_1 = -13.6$ eV.

NEET Trap

What the negative sign means

A free electron at rest, infinitely far from the nucleus ($n = \infty$), is assigned zero energy. A bound electron has lower energy than this, hence the negative sign. As $n$ decreases, $E_n$ grows in magnitude and becomes more negative — the electron is more tightly bound.

$E_\infty = 0$; most negative energy at $n = 1$ → ground state is the most stable.

n = 1 (−13.6 eV) n = 2 (−3.40 eV) n = 3 n = 4 n = 5 n = ∞ (0 eV) Lyman (UV) Balmer (visible) Paschen (IR)
Figure 2. Energy-level diagram for hydrogen. Levels crowd together near $E = 0$ because $E_n = -13.6/n^2$ eV. Each downward jump emits a photon; transitions ending at $n = 1$ form the Lyman series, at $n = 2$ the Balmer series, and at $n = 3$ the Paschen series.

Hydrogen-like Ions

Bohr's theory works for any species that has just one electron — the so-called hydrogen-like ions such as $\ce{He+}$, $\ce{Li^2+}$ and $\ce{Be^3+}$. The energy and radius then depend on the atomic number $Z$:

$$E_n = -2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right)\ \text{J}, \qquad r_n = 52.9\left(\frac{n^2}{Z}\right)\ \text{pm}.$$

As $Z$ increases, the energy becomes more negative and the radius becomes smaller, so the electron is bound more tightly to the more highly charged nucleus. The velocity of the orbiting electron, by contrast, increases with $Z$ and decreases with $n$.

SpeciesZGround-state energy E₁First-orbit radius r₁
$\ce{H}$1−2.18 × 10⁻¹⁸ J52.9 pm
$\ce{He+}$2−8.72 × 10⁻¹⁸ J26.45 pm
$\ce{Li^2+}$3−19.62 × 10⁻¹⁸ J17.63 pm
Keep going

Bohr's fixed orbits are eventually replaced by probability clouds — see towards the quantum mechanical model.

Explaining the Line Spectrum

The triumph of the model is that it explains the hydrogen line spectrum quantitatively. Energy is absorbed when the electron jumps from a lower orbit to a higher one, and emitted when it falls from a higher orbit to a lower one. The gap between two orbits is

$$\Delta E = E_f - E_i = R_H\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) = 2.18 \times 10^{-18}\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\ \text{J}.$$

By the Bohr frequency rule the photon's frequency is $\nu = \Delta E / h$, which works out as

$$\nu = 3.29 \times 10^{15}\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\ \text{Hz}.$$

Dividing by the speed of light gives the wavenumber form, which is exactly the empirical formula Johannes Rydberg had fitted to the data years earlier:

$$\bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\ \text{m}^{-1}.$$

The value $109677\ \text{cm}^{-1}$ is the Rydberg constant for hydrogen. Because the right-hand side can take only specific values (the $n$'s are integers), only specific photon energies are possible — and those are exactly the sharp lines that are observed. For absorption $n_f > n_i$ and the bracket is positive; for emission $n_i > n_f$ and $\Delta E$ is negative, meaning energy is released.

The Spectral Series

The lines group themselves into series, each defined by the orbit $n_f$ at which the electron lands. Of all the elements, hydrogen has the simplest line spectrum; the Balmer series (landing on $n = 2$) is the only one in the visible region.

SeriesFinal orbit n_fStarting orbits n_iSpectral region
Lyman12, 3, 4, …Ultraviolet
Balmer23, 4, 5, …Visible
Paschen34, 5, 6, …Infrared
Brackett45, 6, 7, …Infrared
Pfund56, 7, 8, …Infrared
NEET Trap

Which series is visible?

Only the Balmer series (transitions ending at $n = 2$) lies in the visible region. Lyman is ultraviolet; Paschen, Brackett and Pfund are all infrared. A common error is to call Lyman "visible" because it has the largest energy gaps — large energy means UV, not visible.

Order of energy gap: Lyman > Balmer > Paschen, i.e. UV > visible > IR.

Worked Examples

Example 1 · Energy of a level

Find the energy of the electron in the $n = 2$ state of the hydrogen atom, and the energy released when the electron falls from $n = 5$ to $n = 2$.

Using $E_n = -2.18 \times 10^{-18}/n^2$ J:

$$E_2 = -\frac{2.18 \times 10^{-18}}{2^2} = -0.545 \times 10^{-18}\ \text{J} = -5.45 \times 10^{-19}\ \text{J}.$$

For the $5 \to 2$ emission (a Balmer line):

$$\Delta E = 2.18 \times 10^{-18}\left(\frac{1}{5^2} - \frac{1}{2^2}\right) = 2.18 \times 10^{-18}(0.04 - 0.25) = -4.58 \times 10^{-19}\ \text{J}.$$

The magnitude $4.58 \times 10^{-19}$ J is emitted. The corresponding frequency is $\nu = \Delta E/h = (4.58 \times 10^{-19})/(6.626 \times 10^{-34}) \approx 6.91 \times 10^{14}$ Hz, a line in the visible Balmer series.

Example 2 · Wavelength of a Balmer line

Calculate the wavelength of the $\ce{H}$ emission line for the $n = 3 \to n = 2$ transition (the first Balmer line, H-alpha).

Use the wavenumber form with $n_f = 2$, $n_i = 3$:

$$\bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.09677 \times 10^{7}\left(\frac{1}{4} - \frac{1}{9}\right)\ \text{m}^{-1}.$$

$$\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} = 0.13889.$$

$$\bar{\nu} = 1.09677 \times 10^{7} \times 0.13889 = 1.5233 \times 10^{6}\ \text{m}^{-1}.$$

Therefore $\lambda = 1/\bar{\nu} = 6.565 \times 10^{-7}\ \text{m} = 656.5$ nm — the familiar red H-alpha line.

Example 3 · A hydrogen-like ion

Calculate the energy and the radius of the first orbit of $\ce{He+}$.

For $\ce{He+}$, $Z = 2$ and $n = 1$. Using $E_n = -2.18 \times 10^{-18}(Z^2/n^2)$ J:

$$E_1 = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} = -8.72 \times 10^{-18}\ \text{J}.$$

Using $r_n = 52.9\,(n^2/Z)$ pm:

$$r_1 = 52.9 \times \frac{1^2}{2} = 26.45\ \text{pm} = 0.02645\ \text{nm}.$$

Limitations of the Model

Bohr's model was a major advance over Rutherford's, accounting for the stability and the line spectrum of hydrogen and hydrogen-like ions. Yet it is too simple to survive closer scrutiny.

ShortcomingWhat it fails to explain
Fine structureThe closely spaced doublet lines seen with high-resolution spectroscopy.
Multi-electron atomsThe spectrum of any atom with more than one electron — even neutral helium ($\ce{He}$), which has two electrons.
Zeeman effectThe splitting of spectral lines in a magnetic field.
Stark effectThe splitting of spectral lines in an electric field.
Chemical bondingThe ability of atoms to combine into molecules through chemical bonds.

These failures stem from a deeper problem: the model treats the electron as a particle on a definite circular path, which violates the wave nature of matter and the Heisenberg uncertainty principle. Recognising this drove physicists toward a fully quantum mechanical description of the atom.

Quick Recap

Bohr's Model in One Glance

  • Four postulates: stationary orbits, constant energy, frequency rule $\nu = \Delta E/h$, and quantised angular momentum $m_e v r = nh/2\pi$.
  • Radius: $r_n = 52.9\,(n^2/Z)$ pm; first Bohr orbit of hydrogen = 52.9 pm.
  • Energy: $E_n = -2.18 \times 10^{-18}\,(Z^2/n^2)$ J $= -13.6\,(Z^2/n^2)$ eV; negative because the bound electron lies below the free-electron zero.
  • Transition: $\Delta E = R_H\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$, reproducing the Rydberg formula $\bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ m⁻¹.
  • Series by final orbit: Lyman (n = 1, UV), Balmer (n = 2, visible), Paschen/Brackett/Pfund (n = 3,4,5, IR).
  • Valid only for one-electron species; fails on fine structure, multi-electron spectra, Zeeman/Stark effects and bonding.

NEET PYQ Snapshot — Bohr's Model for Hydrogen Atom

Radius, energy and spectral-series calculations are repeat performers; here are recent NEET questions drawn straight from this subtopic.

NEET 2025 · Q.48

The ratio of the wavelengths of light absorbed by a hydrogen atom for the $n = 2 \to n = 3$ and $n = 4 \to n = 6$ transitions, respectively, is:

  • (1) 1 : 4
  • (2) 36 : 1
  • (3) 1 : 16
  • (4) 1 : 9
Answer: (1) 1 : 4

$\Delta E_{2\to3} = R_H(\tfrac14 - \tfrac19) = \tfrac{5}{36}R_H$ and $\Delta E_{4\to6} = R_H(\tfrac{1}{16} - \tfrac{1}{36}) = \tfrac{20}{576}R_H = \tfrac{5}{144}R_H$. Since $\lambda \propto 1/\Delta E$, the ratio $\lambda_{2\to3}:\lambda_{4\to6} = \Delta E_{4\to6}:\Delta E_{2\to3} = \tfrac{5}{144}:\tfrac{5}{36} = 1:4$.

NEET 2025 · Q.83

Energy and radius of the first Bohr orbit of $\ce{He+}$ and $\ce{Li^2+}$ are [Given $R_H = 2.18 \times 10^{-18}$ J, $a_0 = 52.9$ pm]:

  • (1) $E(\ce{Li^2+}) = -8.72 \times 10^{-16}$ J, $r = 17.6$ pm; $E(\ce{He+}) = -19.62 \times 10^{-16}$ J, $r = 17.6$ pm
  • (2) $E(\ce{Li^2+}) = -19.62 \times 10^{-18}$ J, $r = 17.6$ pm; $E(\ce{He+}) = -8.72 \times 10^{-18}$ J, $r = 26.4$ pm
  • (3) $E(\ce{Li^2+}) = -8.72 \times 10^{-18}$ J, $r = 26.4$ pm; $E(\ce{He+}) = -19.62 \times 10^{-18}$ J, $r = 17.6$ pm
  • (4) $E(\ce{Li^2+}) = -19.62 \times 10^{-16}$ J, $r = 17.6$ pm; $E(\ce{He+}) = -8.72 \times 10^{-16}$ J, $r = 26.4$ pm
Answer: (2)

$E_n = -2.18 \times 10^{-18}(Z^2/n^2)$ J. For $\ce{He+}$ ($Z=2$): $E = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18}$ J, $r = 52.9/2 = 26.4$ pm. For $\ce{Li^2+}$ ($Z=3$): $E = -2.18 \times 10^{-18} \times 9 = -19.62 \times 10^{-18}$ J, $r = 52.9/3 = 17.6$ pm.

NEET 2024 · Q.68

The energy of an electron in the ground state ($n = 1$) for the $\ce{He+}$ ion is $-x$ J. Then the energy of an electron in the $n = 2$ state for the $\ce{Be^3+}$ ion (in J) is:

  • (1) $-x$
  • (2) $-x/9$
  • (3) $-4x$
  • (4) $-4x/9$
Answer: (1) −x

$E \propto Z^2/n^2$. For $\ce{He+}$: $Z^2/n^2 = 4/1 = 4$, giving $-x$. For $\ce{Be^3+}$ at $n = 2$: $Z^2/n^2 = 16/4 = 4$. The two are equal, so the energy is again $-x$.

NEET 2022 · Q.97

If the radius of the second Bohr orbit of the $\ce{He+}$ ion is 105.8 pm, what is the radius of the third Bohr orbit of the $\ce{Li^2+}$ ion?

  • (1) 15.87 pm
  • (2) 1.587 pm
  • (3) 158.7 Å
  • (4) 158.7 pm
Answer: (4) 158.7 pm

$r_n = 52.9\,(n^2/Z)$ pm, so $r \propto n^2/Z$. Ratio: $\dfrac{r_3(\ce{Li^2+})}{r_2(\ce{He+})} = \dfrac{3^2/3}{2^2/2} = \dfrac{3}{2}$. Hence $r_3(\ce{Li^2+}) = 105.8 \times \tfrac{3}{2} = 158.7$ pm.

FAQs — Bohr's Model for Hydrogen Atom

Quick answers to the questions NEET aspirants ask most about Bohr's model.

What is the radius of the first Bohr orbit of the hydrogen atom?

The radius of the first stationary state of hydrogen (n = 1, Z = 1), called the Bohr orbit, is 52.9 pm. The radii of higher orbits follow rₙ = 52.9 · (n²/Z) pm, so the radius grows as n² and shrinks as Z increases. For hydrogen the second orbit lies at 4 × 52.9 = 211.6 pm and the third at 9 × 52.9 = 476.1 pm.

Why is the energy of an electron in a Bohr orbit negative?

The negative sign means the energy of the bound electron is lower than that of a free electron at rest, which is taken as the zero of energy at n = ∞. As the electron moves closer to the nucleus (n decreases), Eₙ becomes larger in magnitude and more negative. The most negative value, Eₙ = −2.18×10⁻¹⁸ J at n = 1, is the most stable ground state.

How does Bohr's model explain the hydrogen line spectrum?

Energy is absorbed when the electron jumps from a lower orbit to a higher one and emitted when it falls from a higher orbit to a lower one. Each jump corresponds to a fixed energy gap ΔE = Ef − Ei, so only photons of certain discrete frequencies (ν = ΔE/h) are involved. These discrete frequencies appear as sharp lines, giving the observed line spectrum rather than a continuous band.

What are the Lyman, Balmer and Paschen series in the hydrogen spectrum?

They are series of lines named after the final orbit to which the electron falls. Lyman series ends at n = 1 and lies in the ultraviolet; Balmer series ends at n = 2 and is the only series in the visible region; Paschen series ends at n = 3 and lies in the infrared. The Brackett (n = 4) and Pfund (n = 5) series also lie in the infrared.

Can Bohr's model be applied to ions other than hydrogen?

Yes, but only to hydrogen-like (one-electron) species such as He⁺, Li²⁺ and Be³⁺. For these the energy is Eₙ = −2.18×10⁻¹⁸ · (Z²/n²) J and the radius is rₙ = 52.9 · (n²/Z) pm. The model fails for any atom or ion with two or more electrons, including the neutral helium atom.

What are the main limitations of Bohr's model?

Bohr's model cannot explain the fine structure (closely spaced doublet lines) of the hydrogen spectrum, nor the spectra of multi-electron atoms such as helium. It fails to account for the Zeeman effect (line splitting in a magnetic field) and the Stark effect (splitting in an electric field), and it cannot explain how atoms form chemical bonds. These shortcomings drove the development of the quantum mechanical model.