Chapter 12 · Atoms

12.2 Atomic Spectra and Bohr's Model of the Hydrogen Atom

Every element sends out light of only certain sharp wavelengths. For hydrogen they obey one simple formula, and Bohr's three postulates explain it — fixing the radius, the speed and the energy of every allowed orbit.

Point a telescope fitted with a spectrograph at a glowing cloud of gas in the Milky Way, such as the Orion Nebula, and its light splits into a few bright coloured lines — the strongest a deep red line at $656$ nm. Astronomers recognise it at once: it is hydrogen. The same red line, shifted to longer wavelengths, appears in the light of galaxies billions of light-years away, and the size of the shift tells how fast they are receding. In 1868 a yellow line in the spectrum of the Sun's edge matched no element known on Earth; it was named helium, after the Greek word for the Sun, and found on Earth only in 1895. Every element has its own set of lines, and spectroscopy — reading those lines — is how we know the composition of stars, of flames in a steel furnace and of pollutants in city air.

Why only certain wavelengths? In this section we first describe the facts: emission and absorption line spectra, and the remarkably simple formula that gives every line of hydrogen. Then we follow Niels Bohr, who in 1913 combined Rutherford's nuclear atom (Section 12.1) with Planck's and Einstein's idea that light energy comes in quanta (Chapter 11). His three postulates fix the radius, speed and energy of the electron's orbits and — as Section 12.3 shows — reproduce the hydrogen formula exactly.

Emission and Absorption Spectra

A hot solid, such as the filament of an old light bulb or a red-hot iron bar, emits light of all wavelengths; passed through a prism or a grating it gives a continuous spectrum, a band of colour with no gaps. A gas of separate atoms behaves quite differently.

📘 Definition 12.5 — Emission line spectrum
When a gas or vapour of an element is excited — by an electric discharge through it at low pressure, or by heating in a flame — its atoms emit light of only certain sharp, discrete wavelengths. The pattern of bright lines on a dark background is the emission line spectrum of the element. No two elements have the same pattern.
📘 Definition 12.6 — Absorption line spectrum
When white light (a continuous spectrum) passes through a gas, the atoms absorb certain wavelengths, which then appear as dark lines in the continuous spectrum: the absorption line spectrum. The dark lines of an element are at wavelengths where that element also has bright emission lines (although not every emission line need appear in absorption).

The dark lines in sunlight, first mapped carefully by Joseph von Fraunhofer in 1814, are absorption lines formed as light from the Sun's hot interior passes through its cooler outer layers. Matching them with laboratory emission lines identified hydrogen, sodium, iron, calcium and dozens of other elements in the Sun. You can see all three kinds of spectrum in the teaser on the chapter page.

The Spectral Series of Hydrogen

Hydrogen, the simplest atom, has the simplest spectrum. In the visible region it has four bright lines, at about $656$, $486$, $434$ and $410$ nm, with more lines crowding together towards $365$ nm in the near ultraviolet. In 1885 a Swiss schoolteacher, Johann Balmer, found by trial that their wavelengths fit a simple formula. Written in the form that Johannes Rydberg later generalised, $$\frac1\lambda=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right),\qquad n=3,4,5,\dots$$ where $R=1.097\times10^7\ \text{m}^{-1}$ is a constant. The set of lines is called the Balmer series. Similar series were later found in the ultraviolet and infrared.

📐 Result 12.4 — The spectral series of hydrogen (Rydberg formula)

Every line in the spectrum of atomic hydrogen has a wavelength $\lambda$ given by $$\frac1\lambda=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right),$$ where $n_f$ and $n_i$ are positive whole numbers with $n_i\gt n_f$, and $R=1.097\times10^7\ \text{m}^{-1}$ is the Rydberg constant. Lines with the same $n_f$ form a series:

  • Lyman series: $n_f=1$, $n_i=2,3,4,\dots$ — ultraviolet;
  • Balmer series: $n_f=2$, $n_i=3,4,5,\dots$ — visible and near ultraviolet;
  • Paschen series: $n_f=3$, $n_i=4,5,6,\dots$ — infrared;
  • Brackett series ($n_f=4$) and Pfund series ($n_f=5$) — further infrared.

As $n_i\to\infty$ the lines of a series crowd together towards the series limit, $\lambda_\infty=\dfrac{n_f^2}{R}$, the shortest wavelength of the series. The longest wavelength of a series is its first line, $n_i=n_f+1$.

The wavenumber $1/\lambda$ is always the difference of two terms of the form $R/n^2$. That pattern — every line is a difference of two numbers belonging to the atom — is the strongest hint that the atom has a set of allowed energies, and that a line is emitted when the atom drops from one to another. Bohr's model makes the hint precise.

✏️ Example 1 — The visible lines of hydrogen
Using the Rydberg formula, find (a) the wavelength of the first line of the Balmer series (called H-alpha), (b) the wavelength of the second line (H-beta), and (c) the series limit. (d) What colours are the first two lines?
(a) n = 3 $\dfrac1\lambda=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}$, so $\lambda=\dfrac{36}{5R}=\dfrac{36}{5(1.097\times10^7)}$ $=6.563\times10^{-7}$ m $\approx656.3$ nm.
(b) n = 4 $\dfrac1\lambda=R\left(\dfrac14-\dfrac1{16}\right)=\dfrac{3R}{16}$, so $\lambda=\dfrac{16}{3R}\approx486.2$ nm.
(c) Limit $n_i\to\infty$: $\lambda_\infty=\dfrac4R=\dfrac{4}{1.097\times10^7}$ $\approx3.646\times10^{-7}$ m $=364.6$ nm, in the near ultraviolet.
(d) Colours $656$ nm is red and $486$ nm is blue-green (cyan) — the two strongest lines that give a hydrogen lamp its pinkish glow.
Check The ratio of longest to shortest Balmer wavelength is $\dfrac{36/5}{4}=1.8$: all Balmer lines lie between $365$ and $656$ nm. Interactive 1 lists the same values.
✏️ Example 2 — The range of the Lyman and Paschen series
Find the longest and the shortest wavelengths of (a) the Lyman series and (b) the Paschen series. In which part of the electromagnetic spectrum does each series lie, and do the two series overlap?
(a) Lyman Longest ($2\to1$): $\lambda=\dfrac{1}{R(1-\frac14)}=\dfrac{4}{3R}\approx121.5$ nm. Shortest (limit): $\lambda=\dfrac1R\approx91.2$ nm. Entirely in the ultraviolet (Section 8.2).
(b) Paschen Longest ($4\to3$): $\dfrac1\lambda=R\left(\dfrac19-\dfrac1{16}\right)=\dfrac{7R}{144}$, $\lambda=\dfrac{144}{7R}\approx1875$ nm. Shortest: $\lambda=\dfrac9R\approx820.4$ nm. Both in the infrared.
Overlap? Lyman: $91$–$122$ nm; Balmer: $365$–$656$ nm; Paschen: $820$–$1875$ nm. These three do not overlap. (The Brackett series, $1458$–$4051$ nm, does overlap Paschen.)
✏️ Example 3 — Identifying a line
An ultraviolet line of hydrogen is measured at $102.6$ nm. Show that it belongs to the Lyman series and find $n_i$.
Series $102.6$ nm lies between $91.2$ and $121.5$ nm, the range of the Lyman series ($n_f=1$); no other series reaches below $364$ nm.
Solve $\dfrac{1}{\lambda R}=1-\dfrac1{n_i^2}$: $\dfrac{1}{(1.026\times10^{-7})(1.097\times10^7)}=0.8885$, so $\dfrac{1}{n_i^2}=0.1115$ and $n_i^2\approx8.97$, $n_i=3$.
Check $\lambda(3\to1)=\dfrac{9}{8R}\approx102.6$ nm ✓. This is the second line of the Lyman series (Lyman-beta).
Interactive 1 — The Hydrogen Spectrum, Series by Series

Every line here is computed from the Rydberg formula, $\tfrac1\lambda=R\left(\tfrac{1}{n_f^2}-\tfrac{1}{n_i^2}\right)$ with $R=1.097\times10^7\ \text{m}^{-1}$, and placed on a logarithmic wavelength axis from $80$ nm to $8000$ nm. Each series has its own row; the bottom strip shows them all together, with the visible band coloured. Choose a series and an upper number $n_i$ (or press Play) to watch the lines march towards the series limit.

3
transition—
wavelength $\lambda$—
region—
series limit—

Bohr's Postulates

Rutherford's atom, with electrons orbiting a tiny nucleus, was right about the structure of the atom, but classical physics predicted that it should collapse in about $10^{-11}$ s while radiating a continuous spectrum (Section 12.1). Real atoms are stable and emit sharp lines. Bohr kept the nuclear atom and Newton's mechanics for the orbits, but added three new rules that broke with classical physics where they had to.

📐 Principle 12.5 — Bohr's postulates (hydrogen atom)

1. Stationary orbits. The electron moves round the nucleus in circular orbits under the Coulomb attraction, obeying Newton's laws — but only certain orbits are allowed. In these stationary orbits the electron does not radiate, even though it is accelerating, so its energy stays constant.

2. Quantisation of angular momentum. The allowed orbits are those for which the angular momentum of the electron is a whole-number multiple of $h/2\pi$: $$L=mvr=\frac{nh}{2\pi},\qquad n=1,2,3,\dots$$ The whole number $n$ is the principal quantum number of the orbit.

3. Frequency condition. The atom emits (or absorbs) radiation only when the electron jumps from one stationary orbit to another. A single photon is emitted (or absorbed), whose energy equals the difference of the energies of the two orbits: $$h\nu=E_i-E_f,$$ where $E_i$ is the energy of the initial orbit and $E_f$ that of the final one.

The first postulate simply declares that the classical prediction of radiation does not apply to stationary orbits; the third says how light is emitted instead. The second picks out which orbits are allowed. It looked arbitrary in 1913 — Bohr chose it because it gave the right answer — but de Broglie's matter waves later explained it (Section 12.3). Notice that the frequency $\nu$ of the emitted light has nothing to do with the frequency of revolution of the electron, in sharp contrast to classical physics.

Radius, Speed and Energy of the Bohr Orbits

Now combine Newton's second law with Bohr's second postulate. For an electron of mass $m$ and charge $-e$ in an orbit of radius $r$ about a proton:

Newton's second law (Coulomb force provides the centripetal force): $$\frac{mv^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}.\qquad(1)$$ Bohr's quantum condition: $$mvr=\frac{nh}{2\pi}.\qquad(2)$$

Speed. Multiply (1) by $r^2$ to get $mv^2r=\dfrac{e^2}{4\pi\varepsilon_0}$, i.e. $v\,(mvr)=\dfrac{e^2}{4\pi\varepsilon_0}$. Substituting $mvr$ from (2), $$v_n=\frac{e^2}{4\pi\varepsilon_0}\cdot\frac{2\pi}{nh}=\frac{e^2}{2\varepsilon_0nh}.$$

Radius. From (2), $r=\dfrac{nh}{2\pi mv_n}$; inserting $v_n$, $$r_n=\frac{nh}{2\pi m}\cdot\frac{2\varepsilon_0nh}{e^2}=\frac{\varepsilon_0h^2}{\pi me^2}\,n^2.$$

Energy. By Result 12.3, $E=-\dfrac{e^2}{8\pi\varepsilon_0r}$ for any circular orbit; with $r=r_n$, $$E_n=-\frac{e^2}{8\pi\varepsilon_0}\cdot\frac{\pi me^2}{\varepsilon_0h^2n^2}=-\frac{me^4}{8\varepsilon_0^2h^2}\cdot\frac{1}{n^2}.$$

📐 Result 12.6 — The Bohr orbits of hydrogen

In the $n$th stationary orbit of the hydrogen atom ($n=1,2,3,\dots$):

$$r_n=\frac{\varepsilon_0h^2}{\pi me^2}\,n^2=a_0n^2,\qquad a_0\approx0.53\ \text{Å}=5.3\times10^{-11}\ \text{m};$$ $$v_n=\frac{e^2}{2\varepsilon_0h}\cdot\frac1n\approx\frac{2.18\times10^6}{n}\ \text{m/s};$$ $$E_n=-\frac{me^4}{8\varepsilon_0^2h^2}\cdot\frac{1}{n^2}\approx-\frac{13.6}{n^2}\ \text{eV}.$$

$a_0$ is the Bohr radius, the radius of the smallest orbit. The kinetic energy is $K_n=-E_n=\dfrac{13.6}{n^2}$ eV and the potential energy is $U_n=2E_n=-\dfrac{27.2}{n^2}$ eV. Thus $r\propto n^2$, $v\propto1/n$ and $E\propto-1/n^2$.

With the course constants, $a_0=\dfrac{(8.85\times10^{-12})(6.63\times10^{-34})^2}{\pi(9.1\times10^{-31})(1.6\times10^{-19})^2}$ $\approx5.3\times10^{-11}$ m and $\dfrac{me^4}{8\varepsilon_0^2h^2}\approx2.17\times10^{-18}$ J $\approx13.5$ eV; more precise values of the constants give $13.6$ eV, which agrees with the measured energy needed to ionise a hydrogen atom. That agreement — from nothing but $m$, $e$, $h$ and $\varepsilon_0$ — was the first triumph of the model. The speed in the first orbit is about $c/137$, so non-relativistic mechanics is a fair approximation.

✏️ Example 4 — The second and third orbits
For the electron of a hydrogen atom in the orbits $n=2$ and $n=3$, find the radius, the speed and the total energy. Also find the kinetic and potential energies for $n=2$.
Scaling $r_n=0.53n^2$ Å, $v_n=\dfrac{2.18\times10^6}{n}$ m/s, $E_n=-\dfrac{13.6}{n^2}$ eV.
n = 2 $r_2=0.53\times4=2.12$ Å; $v_2=1.09\times10^6$ m/s; $E_2=-\dfrac{13.6}{4}=-3.4$ eV.
n = 3 $r_3=0.53\times9=4.77$ Å; $v_3\approx7.3\times10^5$ m/s; $E_3=-\dfrac{13.6}{9}\approx-1.51$ eV.
n = 2 energies $K_2=-E_2=+3.4$ eV; $U_2=2E_2=-6.8$ eV.
Check Interactive 2 at $n=2$ and $n=3$ shows these values. Higher orbits are bigger, slower and less tightly bound (energy closer to zero).
✏️ Example 5 — The orbiting electron as a current
For the electron in the ground orbit ($n=1$) of hydrogen, find (a) the time for one revolution, (b) the average current that the circulating charge represents, and (c) the magnetic field this current produces at the nucleus (use $\mu_0=4\pi\times10^{-7}$ T m/A).
(a) Period $T_1=\dfrac{2\pi r_1}{v_1}=\dfrac{2\pi(5.3\times10^{-11})}{2.18\times10^6}$ $\approx1.53\times10^{-16}$ s.
(b) Current A charge $e$ passes any point once per revolution: $I=\dfrac{e}{T_1}=\dfrac{1.6\times10^{-19}}{1.53\times10^{-16}}$ $\approx1.05\times10^{-3}$ A $\approx1.0$ mA.
(c) Field Field at the centre of a circular loop (Section 4.2): $B=\dfrac{\mu_0I}{2r_1}$ $=\dfrac{(4\pi\times10^{-7})(1.05\times10^{-3})}{2(5.3\times10^{-11})}$ $\approx12$ T.
Meaning A milliampere is a large current for one electron, and $12$ T is stronger than the field of most laboratory magnets — atoms are tiny, and fields inside them are huge.
✏️ Example 6 — How the orbit changes from n = 1 to n = 3
An electron in a hydrogen atom moves from the orbit $n=1$ to the orbit $n=3$. By what factor does each of these change: (a) the radius, (b) the speed, (c) the angular momentum, (d) the total energy, (e) the period of revolution?
(a), (b) $r\propto n^2$: $\times9$. $v\propto1/n$: $\times\tfrac13$.
(c) $L=\dfrac{nh}{2\pi}\propto n$: $\times3$.
(d) $E\propto-1/n^2$: $\times\tfrac19$, from $-13.6$ eV to $-1.51$ eV (the energy increases, because it becomes less negative).
(e) $T=\dfrac{2\pi r}{v}\propto\dfrac{n^2}{1/n}=n^3$: $\times27$.
Check $L=mvr\propto\dfrac1n\cdot n^2=n$ ✓, consistent with (c).
Interactive 2 — Bohr Model Explorer

Choose the principal quantum number $n$. On the left the allowed orbits $n=1$ to $6$ are drawn to scale ($r_n=0.53\,n^2$ Å) and the electron circles in the chosen one; on the right the energy of each orbit is marked on a linear scale from $-13.6$ eV to $0$. The readouts are calculated from Result 12.6. (Revolution rates in the drawing are slowed down enormously, and the outer orbits are speeded up relative to the inner ones so that you can see them move; the real period grows as $n^3$.)

2
radius $r_n$—
speed $v_n$—
energy $E_n$—
$K_n$ and $U_n$—
angular momentum $L$—
period $T_n$—
⚠️ Common mistakes with spectra and Bohr orbits

1. Mixing up $n_i$ and $n_f$. In emission the electron falls from $n_i$ (higher) to $n_f$ (lower); $n_f$ names the series. Putting them the other way round gives a negative $1/\lambda$.

2. "The series limit is the longest wavelength." It is the shortest ($n_i\to\infty$); the first line ($n_i=n_f+1$) is the longest.

3. Wrong powers of $n$. $r\propto n^2$, $v\propto1/n$, $E\propto1/n^2$, $L\propto n$, $T\propto n^3$. Write them down before any ratio problem.

4. Sign of the energy. $E_n$ is negative; a higher orbit has a higher (less negative) energy although $|E_n|$ is smaller. The kinetic energy is positive, $K_n=|E_n|$.

5. Confusing the photon frequency with the orbital frequency. In Bohr's model the light emitted has $\nu=(E_i-E_f)/h$, not the electron's frequency of revolution.

🚀 Beyond the board syllabus (JEE Advanced) — hydrogen-like ions

In JEE Advanced 2026 only (it lists Bohr's theory of hydrogen-like atoms); not in the CBSE, JEE Main or NEET 2026 syllabi, which name only the hydrogen atom.

Bohr's derivation works for any ion with a single electron and a nucleus of charge $+Ze$ — He+ ($Z=2$), Li2+ ($Z=3$) and so on. Replacing $e^2$ by $Ze^2$ in the Coulomb force gives $$r_n=\frac{0.53\,n^2}{Z}\ \text{Å},\qquad v_n=\frac{2.18\times10^6\,Z}{n}\ \text{m/s},\qquad E_n=-\frac{13.6\,Z^2}{n^2}\ \text{eV},$$ and the Rydberg formula becomes $\dfrac1\lambda=RZ^2\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$. So the ground-state energy of He+ is $-54.4$ eV and of Li2+ is $-122.4$ eV; the first orbit of He+ has radius $0.265$ Å. A favourite JEE observation: the $4\to2$ transition in He+ releases $13.6\times4\left(\tfrac14-\tfrac1{16}\right)=10.2$ eV, exactly the energy of the $2\to1$ (Lyman-alpha) line of hydrogen. These results are not part of the board syllabus.

Exercises

I
Line spectra and the series of hydrogen. $R=1.097\times10^7\ \text{m}^{-1}$, $c=3.0\times10^8$ m/s.
  • 1 Distinguish between an emission line spectrum and an absorption line spectrum. How are the two related for a given element?
  • 2 Explain how the dark lines in the spectrum of sunlight reveal the elements present in the Sun's outer layers.
  • 3 Find the wavelength of the third line of the Balmer series ($n_i=5$).
  • 4 Find the series limit of the Balmer series.
  • 5 Find the longest wavelength in the Lyman series. In which region does it lie?
  • 6 Find the shortest wavelength in the Paschen series.
  • 7 A hydrogen line is observed at $1282$ nm. To which series does it belong, and what is $n_i$?
  • 8 Find the frequency of the H-alpha line ($656.3$ nm).
1 Emission: bright lines on a dark background from an excited gas. Absorption: dark lines in a continuous spectrum, where a gas has absorbed light. For a given element the dark lines lie at wavelengths of its emission lines.
2 Light from the hot interior passes through the cooler outer gas, whose atoms absorb their own characteristic wavelengths. Each set of dark lines matches the laboratory emission lines of a particular element, which must therefore be present.
3 $\dfrac1\lambda=R\left(\dfrac14-\dfrac1{25}\right)=\dfrac{21R}{100}$; $\lambda=\dfrac{100}{21R}\approx434.1$ nm (violet).
4 $\lambda_\infty=\dfrac4R\approx364.6$ nm.
5 $\lambda=\dfrac{4}{3R}\approx121.5$ nm, ultraviolet.
6 $\lambda_\infty=\dfrac9R\approx820.4$ nm (infrared).
7 $1282$ nm lies in the Paschen range ($820$–$1875$ nm). $\dfrac{1}{\lambda R}=0.0711=\dfrac19-\dfrac1{n_i^2}$ gives $\dfrac1{n_i^2}=0.0400$, $n_i=5$. Check: $\dfrac{225}{16R}\approx1282$ nm ✓.
8 $\nu=\dfrac c\lambda=\dfrac{3.0\times10^8}{6.563\times10^{-7}}\approx4.57\times10^{14}$ Hz.
II
Bohr's postulates. $h=6.63\times10^{-34}$ J s.
  • 1 State Bohr's three postulates for the hydrogen atom.
  • 2 Find the angular momentum of the electron in the orbit $n=3$.
  • 3 Can an electron in a hydrogen atom have angular momentum $2.5\,h/2\pi$? Explain.
  • 4 Find the ratio of the angular momenta of the electron in the orbits $n=4$ and $n=2$.
  • 5 In what way does Bohr's first postulate contradict classical electromagnetism?
  • 6 The electron of a hydrogen atom has angular momentum $4.22\times10^{-34}$ J s. Which orbit is it in?
  • 7 In a transition, a hydrogen atom emits a photon of energy $1.89$ eV. Find its frequency.
1 (i) The electron moves in certain circular stationary orbits without radiating. (ii) In these orbits $mvr=nh/2\pi$, $n=1,2,3,\dots$ (iii) Radiation is emitted or absorbed only in a jump between orbits, as one photon with $h\nu=E_i-E_f$.
2 $L=\dfrac{3h}{2\pi}=\dfrac{3(6.63\times10^{-34})}{2\pi}\approx3.17\times10^{-34}$ J s.
3 No: by the second postulate $L$ must be a whole-number multiple of $h/2\pi$.
4 $L\propto n$: $4:2=2$.
5 Classically an accelerated charge always radiates; an orbiting electron accelerates (centripetally), so it should lose energy continuously. Bohr postulated that in stationary orbits it does not.
6 $n=\dfrac{2\pi L}{h}=\dfrac{2\pi(4.22\times10^{-34})}{6.63\times10^{-34}}\approx4.0$: the orbit $n=4$.
7 $\nu=\dfrac Eh=\dfrac{1.89\times1.6\times10^{-19}}{6.63\times10^{-34}}\approx4.56\times10^{14}$ Hz — the H-alpha line ($3\to2$) again.
III
Radius, speed and energy. Use $a_0=0.53$ Å, $v_1=2.18\times10^6$ m/s and $E_1=-13.6$ eV.
  • 1 Find the radii of the orbits $n=2$, $3$ and $4$.
  • 2 Find the speed of the electron in the orbit $n=3$.
  • 3 Find the energy of the electron in the orbit $n=4$.
  • 4 Find the kinetic and potential energies of the electron in the orbit $n=2$.
  • 5 Which orbit has a radius of $1.325$ nm?
  • 6 Express the speed of the electron in the first orbit as a fraction of the speed of light.
  • 7 The period of revolution in the first orbit is $1.53\times10^{-16}$ s. Find it for the orbit $n=2$.
  • 8 How much energy is needed to remove the electron completely from the orbit $n=3$?
1 $r_n=0.53n^2$ Å: $2.12$ Å, $4.77$ Å and $8.48$ Å.
2 $v_3=\dfrac{2.18\times10^6}{3}\approx7.3\times10^5$ m/s.
3 $E_4=-\dfrac{13.6}{16}=-0.85$ eV.
4 $E_2=-3.4$ eV, so $K_2=+3.4$ eV and $U_2=-6.8$ eV.
5 $n^2=\dfrac{13.25\ \text{Å}}{0.53\ \text{Å}}=25$, so $n=5$.
6 $\dfrac{v_1}{c}=\dfrac{2.18\times10^6}{3.0\times10^8}\approx7.3\times10^{-3}\approx\dfrac{1}{137}$.
7 $T\propto n^3$: $T_2=8\times1.53\times10^{-16}\approx1.22\times10^{-15}$ s.
8 To reach $E=0$ from $E_3=-1.51$ eV needs $1.51$ eV.
IV
Derivations and applications. $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$, $m=9.1\times10^{-31}$ kg, $e=1.6\times10^{-19}$ C, $h=6.63\times10^{-34}$ J s.
  • 1 Starting from Bohr's postulates, derive the radius of the $n$th orbit of hydrogen and show that $r_n\propto n^2$.
  • 2 Evaluate the Bohr radius $a_0=\dfrac{\varepsilon_0h^2}{\pi me^2}$.
  • 3 Evaluate the ground-state energy $E_1=-\dfrac{me^4}{8\varepsilon_0^2h^2}$ in joules and in eV.
  • 4 Show that $\dfrac{v_n}{c}=\dfrac{e^2}{2\varepsilon_0hc}\cdot\dfrac1n$ and evaluate the constant $\dfrac{e^2}{2\varepsilon_0hc}$.
  • 5 Modelling. Radio astronomers detect hydrogen atoms in interstellar clouds in orbits as high as $n=100$. Find the radius of such an atom and the energy needed to ionise it. Find the frequency of the photon emitted in the jump $n=100\to99$, and say in which part of the spectrum it lies.
  • 6 Find the magnetic field at the nucleus produced by the electron in the orbit $n=2$, and compare it with Example 5. (Hint: how does $B=\mu_0I/2r$ depend on $n$?)
1 $\dfrac{mv^2}{r}=\dfrac{e^2}{4\pi\varepsilon_0r^2}$ and $mvr=\dfrac{nh}{2\pi}$. The first gives $v\,(mvr)=\dfrac{e^2}{4\pi\varepsilon_0}$, so $v_n=\dfrac{e^2}{2\varepsilon_0nh}$; then $r_n=\dfrac{nh}{2\pi mv_n}=\dfrac{\varepsilon_0h^2}{\pi me^2}n^2\propto n^2$.
2 $a_0=\dfrac{(8.85\times10^{-12})(4.396\times10^{-67})}{\pi(9.1\times10^{-31})(2.56\times10^{-38})}$ $\approx5.3\times10^{-11}$ m $=0.53$ Å.
3 $E_1=-\dfrac{(9.1\times10^{-31})(1.6\times10^{-19})^4}{8(8.85\times10^{-12})^2(6.63\times10^{-34})^2}$ $\approx-2.17\times10^{-18}$ J $\approx-13.5$ eV (precise constants give $-13.6$ eV).
4 Divide $v_n=\dfrac{e^2}{2\varepsilon_0nh}$ by $c$. The denominator is $2(8.85\times10^{-12})(6.63\times10^{-34})(3.0\times10^8)$ $\approx3.52\times10^{-36}$, so $\dfrac{e^2}{2\varepsilon_0hc}=\dfrac{2.56\times10^{-38}}{3.52\times10^{-36}}$ $\approx7.3\times10^{-3}\approx\dfrac1{137}$ (a pure number, with no units).
5 $r=0.53\times10^4$ Å $=5.3\times10^{-7}$ m (about half a micrometre — a giant atom); ionisation energy $\dfrac{13.6}{10^4}$ eV $=1.36$ meV. $h\nu=13.6\left(\dfrac{1}{99^2}-\dfrac1{100^2}\right)$ eV $\approx2.76\times10^{-5}$ eV, so $\nu\approx6.7\times10^9$ Hz ($\lambda\approx4.5$ cm): microwaves (radio astronomy).
6 $I=\dfrac{e}{T}\propto\dfrac1{n^3}$ and $r\propto n^2$, so $B\propto\dfrac1{n^5}$: $B_2=\dfrac{12.4}{32}\approx0.39$ T.
🎯 Exam corner — spectra and the Bohr model BoardJEE MainJEE AdvancedNEET

E1. Board · 5 marks State Bohr's postulates for the hydrogen atom. Using them, derive expressions for the radius of the $n$th orbit and the total energy of the electron in it. Hence find the radius and energy for $n=2$.

Postulates (i) Stationary circular orbits, no radiation. (ii) $mvr=\dfrac{nh}{2\pi}$. (iii) $h\nu=E_i-E_f$ for a jump.
Radius $\dfrac{mv^2}{r}=\dfrac{e^2}{4\pi\varepsilon_0r^2}$ with $v=\dfrac{nh}{2\pi mr}$ gives $r_n=\dfrac{\varepsilon_0h^2n^2}{\pi me^2}$.
Energy $K=\dfrac{e^2}{8\pi\varepsilon_0r}$, $U=-\dfrac{e^2}{4\pi\varepsilon_0r}$, so $E=-\dfrac{e^2}{8\pi\varepsilon_0r_n}=-\dfrac{me^4}{8\varepsilon_0^2h^2n^2}=-\dfrac{13.6}{n^2}$ eV.
n = 2 $r_2=4a_0=2.12\times10^{-10}$ m; $E_2=-3.4$ eV.
Exam tip Show the step where $mvr$ is substituted — examiners give a mark for it — and write the sign of $E$ with a sentence on its meaning (bound electron).

E2. JEE Main The ratio of the longest wavelength to the shortest wavelength in the Balmer series of hydrogen is (A) $9/5$ (B) $5/9$ (C) $4/3$ (D) $36/5$.

Concept Longest: $3\to2$, $\lambda=\dfrac{36}{5R}$. Shortest: series limit, $\lambda=\dfrac4R$. Ratio $=\dfrac{36}{5\times4}=\dfrac95$.
Answer (A) $9/5$. Exam tip: (D) is $\lambda_\text{max}R$, not a ratio of wavelengths.

E3. NEET For an electron in any Bohr orbit of the hydrogen atom, the ratio of its kinetic energy to its total energy is (A) $-1$ (B) $2$ (C) $\tfrac12$ (D) $-2$.

Concept $K=-E$ (Result 12.3), so $K/E=-1$.
Answer (A). Exam tip: $U/E=2$ and $U/K=-2$ — options (B) and (D) answer different questions.

E4. JEE Main NEET The speed of the electron in the third Bohr orbit of hydrogen is about (A) $7.3\times10^5$ m/s (B) $2.2\times10^6$ m/s (C) $6.5\times10^6$ m/s (D) $2.4\times10^5$ m/s.

Concept $v_n=\dfrac{2.18\times10^6}{n}$ m/s, so $v_3\approx7.3\times10^5$ m/s.
Answer (A). Exam tip: (C) multiplies by $n$ instead of dividing, and (D) divides by $n^2$ — the radius rule, not the speed rule.

E5. JEE Advanced (One or more correct; hydrogen-like ions — beyond NCERT, see the 🚀 box.) For the single electron of a hydrogen-like ion of atomic number $Z$ in the orbit $n$: (A) $r\propto n^2/Z$ (B) $v\propto Z/n$ (C) $E\propto-Z^2/n^2$ (D) the angular momentum is proportional to $Z$.

Concept Replace $e^2$ by $Ze^2$ in the force law: $v=\dfrac{Ze^2}{2\varepsilon_0nh}$, $r=\dfrac{\varepsilon_0h^2n^2}{\pi mZe^2}$, $E=-\dfrac{mZ^2e^4}{8\varepsilon_0^2h^2n^2}$. (A), (B), (C) true.
(D) False: $L=\dfrac{nh}{2\pi}$ whatever the nucleus — the quantum condition does not involve $Z$.
Answer (A), (B), (C). Exam tip: this topic is outside the NCERT syllabus but common in JEE and NEET papers.

E6. Board · Case study An astronomer photographs the spectrum of a glowing gas cloud and finds a strong red line and a weaker blue-green line, which she identifies as the first two lines of the Balmer series of hydrogen. (i) (2 marks) Calculate the wavelengths of these two lines ($R=1.097\times10^7\ \text{m}^{-1}$). (ii) (1 mark) Which series of hydrogen lies entirely in the ultraviolet? (iii) (1 mark) Towards what wavelength do the lines of the Balmer series converge?

(i) $3\to2$: $\lambda=\dfrac{36}{5R}\approx656$ nm (red); $4\to2$: $\lambda=\dfrac{16}{3R}\approx486$ nm (blue-green).
(ii) The Lyman series ($91$–$122$ nm).
(iii) The series limit $\dfrac4R\approx365$ nm, in the near ultraviolet. Exam tip: state which $n$ values you used — that earns the method mark even if the arithmetic slips.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How accurately can the Rydberg constant be measured from a hydrogen spectrum — and how do astronomers use the same lines to measure how fast distant galaxies are receding?

Why it matters. The Rydberg constant is one of the most precisely known numbers in physics; metrology laboratories measure hydrogen lines with lasers to better than one part in $10^{12}$ to test atomic theory. At the other extreme, large sky surveys record millions of galaxy spectra and read each galaxy's recession speed from the shift of its hydrogen lines — the data from which the expansion of the universe is mapped.

What to do.

  1. A student's spectrometer gives these wavelengths for four Balmer lines (illustrative data): $656.3$, $486.1$, $434.0$ and $410.2$ nm for $n_i=3,4,5,6$. Tabulate $x=1/n_i^2$ and $y=1/\lambda$ (in $\text{m}^{-1}$).
  2. The Rydberg formula predicts $y=\dfrac R4-Rx$, a straight line. Plot $y$ against $x$ and draw the best-fit line (or use a spreadsheet's linear fit).
  3. Find $R$ from the slope and, separately, from the intercept. Do the two agree? Compare with $1.097\times10^7\ \text{m}^{-1}$.
  4. In the spectrum of a distant galaxy the H-alpha line appears at $750$ nm. Astronomers define the redshift $z=\dfrac{\lambda_\text{obs}-\lambda_\text{lab}}{\lambda_\text{lab}}$. Find $z$, and predict where H-beta should appear.
  5. For small $z$, the recession speed is about $cz$. Estimate it, and say whether this "small-$z$" estimate is reliable here.

Tools: Graph paper or a spreadsheet with a linear-fit function.

Hand in: The table, the graph with its fitted line, your two values of $R$ with a comment on agreement, and the redshift calculation.

Going further. Why is the redshift method so powerful? Explain why the ratios of the observed wavelengths of different hydrogen lines are the same as in the laboratory, and how this lets an astronomer recognise the hydrogen pattern even when every line has moved.

Instructor note: $y$ values $1.5237,\ 2.0572,\ 2.3041,\ 2.4378\times10^6\ \text{m}^{-1}$; least-squares slope $\approx-1.097\times10^7\ \text{m}^{-1}$, intercept $\approx2.743\times10^6\ \text{m}^{-1}$, so $R\approx1.097\times10^7\ \text{m}^{-1}$ from either. $z=750/656.3-1\approx0.143$; H-beta at $486.1\times1.143\approx556$ nm. $cz\approx4.3\times10^7$ m/s; at $z\approx0.14$ the simple estimate is only rough (relativistic and cosmological corrections are several per cent). Going further: every wavelength is multiplied by the same factor $1+z$, so ratios are unchanged.

🧪 Self-check quiz

0/5 correct

Q1. Which series of hydrogen lines lies in the visible region?

Q2. Bohr's second postulate says that, in an allowed orbit, the electron's

Q3. The radius of the first Bohr orbit of hydrogen is $0.53$ Å. The radius of the orbit $n=3$ is

Q4. The energy of the electron in the orbit $n=2$ of hydrogen is

Q5. The lines of the Lyman series crowd together towards a series limit at about

← 12.1 Alpha-Particle Scattering and Rutherford's Nuclear Model12.3 The Line Spectra of Hydrogen and de Broglie's Explanation →