Bohr Model of the Hydrogen Atom

Why Rutherford's atom failed

Rutherford's nuclear model pictured the electron orbiting the nucleus like a planet around the sun, the Coulomb attraction supplying the centripetal force. NCERT §12.4 identifies two fatal objections. First, an electron moving in a circle is continuously accelerated, and classical electromagnetic theory requires an accelerating charge to radiate. The electron should therefore lose energy continuously, spiral inward and collapse into the nucleus — so the atom could not be stable.

Second, as the electron spiralled in, its frequency of revolution would change continuously, so the emitted light would span a continuous band of frequencies. This contradicts the observed hydrogen spectrum, which consists of sharp discrete lines. Classical physics, in short, could explain neither the stability of the atom nor its line spectrum.

The three postulates

Bohr kept Rutherford's nucleus but grafted on the new quantum ideas of Planck and Einstein. He accepted that classical electromagnetism does not govern atomic-scale processes, and stated his theory as three postulates (NCERT §12.4).

The second postulate is the engine of the model. With Planck's constant $h = 6.6 \times 10^{-34}$ J s, the quantisation condition is

$$ L = m v_n r_n = \dfrac{nh}{2\pi}, \qquad n = 1, 2, 3, \dots $$

Here $n$ is the principal quantum number. Because $L$ can take only the discrete values $h/2\pi,\, 2h/2\pi,\, 3h/2\pi, \dots$, the allowed radii and energies are also discrete. The third postulate then ties these discrete energies to the discrete frequencies seen in the spectrum, $h\nu = E_i - E_f$ with $E_i > E_f$.

Radius of the nth orbit

The radius is not assumed — it is forced by combining classical mechanics with the second postulate. Start from the dynamical condition that NCERT §12.4 inherits from the Rutherford model: for a stable circular orbit the Coulomb attraction between the nucleus ($+Ze$) and the electron ($-e$) supplies exactly the centripetal force,

$$ \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Ze^2}{r_n^{\,2}} \;=\; \dfrac{m v_n^{\,2}}{r_n} \qquad\Longrightarrow\qquad m v_n^{\,2} \;=\; \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Ze^2}{r_n}. \tag{1} $$

This single equation contains two unknowns, $r_n$ and $v_n$, so classical physics alone cannot fix the orbit — any radius would do. Bohr's second postulate removes the freedom by quantising the angular momentum:

$$ m v_n r_n = \dfrac{nh}{2\pi} \qquad\Longrightarrow\qquad v_n = \dfrac{nh}{2\pi m r_n}. \tag{2} $$

Substituting $v_n$ from (2) into (1) eliminates the velocity. The left side becomes $m\!\left(\dfrac{nh}{2\pi m r_n}\right)^{2} = \dfrac{n^2 h^2}{4\pi^2 m r_n^{\,2}}$, and setting that equal to the right side of (1) gives

$$ \dfrac{n^2 h^2}{4\pi^2 m r_n^{\,2}} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Ze^2}{r_n} \qquad\Longrightarrow\qquad r_n = \dfrac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}. \tag{3} $$

Equation (3) is NCERT Eq. 12.7. Collecting the constants for the ground state of hydrogen ($n = 1$, $Z = 1$) gives the Bohr radius, which NCERT and NIOS §24.2.1 both quote as $a_0 = 5.3 \times 10^{-11}$ m. Writing the result in terms of $a_0$,

$$ r_n = \dfrac{n^2}{Z}\,(0.529\ \text{Å}), \qquad a_0 = 0.529\ \text{Å} = 5.3 \times 10^{-11}\ \text{m}. $$

Because $r_n \propto n^2$ at fixed $Z$, the $n = 2$ and $n = 3$ orbits of hydrogen sit at $4a_0$ and $9a_0$; and because $r_n \propto 1/Z$, a more highly charged nucleus pulls every orbit inward.

Velocity and energy of the electron

The orbital speed follows immediately by feeding the quantised radius (3) back into the quantisation condition (2), $v_n = nh/(2\pi m r_n)$. Substituting $r_n = n^2 h^2 \varepsilon_0 / \pi m Z e^2$ and cancelling one power of $n$ gives

$$ v_n = \dfrac{nh}{2\pi m}\cdot\dfrac{\pi m Z e^2}{n^2 h^2 \varepsilon_0} = \dfrac{Z e^2}{2\varepsilon_0 n h} \qquad\Longrightarrow\qquad v_n \propto \dfrac{Z}{n}. $$

The electron therefore moves fastest in the innermost orbit and in atoms of larger nuclear charge, and slows as $n$ rises. For hydrogen in the ground state ($n = Z = 1$) the constants evaluate to $v_1 \approx 2.2 \times 10^{6}$ m/s, the value NCERT Example 12.3 quotes.

To find the energy, add the kinetic and potential parts. From the force balance (1), the kinetic energy of the electron is

$$ K = \tfrac{1}{2}m v_n^{\,2} = \dfrac{1}{2}\cdot\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Ze^2}{r_n} = +\,\dfrac{Ze^2}{8\pi\varepsilon_0 r_n}. $$

The electrostatic potential energy of the attractive nucleus–electron pair is negative and twice as large in magnitude:

$$ U = -\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Ze^2}{r_n} = -\,\dfrac{Ze^2}{4\pi\varepsilon_0 r_n} = -2K. $$

Hence the total energy $E_n = K + U = K - 2K = -K$ — the relation NCERT Eq. 12.4 records as $E = -\dfrac{Ze^2}{8\pi\varepsilon_0 r_n}$. Inserting the quantised radius (3) for $r_n$ collapses the constants into the headline result (NCERT Eq. 12.8–12.10):

$$ E_n = -\dfrac{m Z^2 e^4}{8\varepsilon_0^{\,2} n^2 h^2} = -\dfrac{13.6\,Z^2}{n^2}\ \text{eV} \qquad\Longrightarrow\qquad E_n \propto -\dfrac{Z^2}{n^2}. $$

Two features deserve emphasis. The energy is negative, signifying that the electron is bound — work must be done on it to reach the $E = 0$ free state; positive energy would mean an unbound electron. And the bookkeeping $K = -E_n$, $U = 2E_n$ holds in every orbit, so the binding energy in any level is numerically equal to its kinetic energy. This $K : E = 1 : -1$ ratio is itself a recurring NEET question.

Energy levels of hydrogen

For hydrogen ($Z = 1$), $E_n = -13.6/n^2$ eV. The energy is most negative in the ground state and rises toward zero as $n$ increases; at $n = \infty$ the energy is $0$ eV, corresponding to the electron removed and at rest (NCERT §12.4.1).

The minimum energy needed to free the ground-state electron is therefore $13.6$ eV — the ionisation energy of hydrogen, in excellent agreement with experiment. The excited levels crowd together as $n$ grows, since the spacing falls off as $1/n^2$. The energy delivered to or absorbed by the atom in any process is always a difference between two of these levels.

Scaling with n and Z

Most NEET problems on the Bohr model reduce to a single skill: reading off how $r_n$, $v_n$ and $E_n$ scale with the quantum number $n$ and the nuclear charge $Z$. The table below collects all three for a hydrogenic atom.

Limitations of the model

Bohr's model brilliantly reproduces the gross features of hydrogen, but NCERT §12.6 lists its boundaries. It applies only to hydrogenic atoms — a nucleus of charge $+Ze$ with a single electron, such as H, He⁺ and Li²⁺. It cannot be extended even to two-electron helium, because it accounts only for the nucleus-electron force and ignores the electron-electron repulsions that appear in multi-electron atoms.

The model also cannot explain the relative intensities of spectral lines — why some transitions are stronger than others. And as the "Points to Ponder" note, the sharp orbital picture conflicts with the uncertainty principle; it was later replaced by full quantum mechanics, in which Bohr's orbits become regions of high probability for finding the electron.