Latest
  • Admissions openClass 11 Science, 2027‑28: JEE, NEET and MHT‑CET with junior college and hostel under one roofApply now
  • IMGSAT 2027Free scholarship and admission test for Class 10 students, every Saturday and Sunday at our Nashik campusRegister
  • Free Foundation 2026‑27Evening classes for Class 10 in Physics, Chemistry, Maths and Biology, taught by our IITian and doctor facultyJoin free

+91 70303 00666

Physics · Class 12 · Chapter 12

Atoms

From Rutherford's scattering experiment to Bohr's quantised orbits, this chapter builds the model that explains the hydrogen spectrum. Almost every question uses En = −13.6/n2 eV in some form, so know the energy level diagram well.

In this chapter: Thomson's model, alpha-particle scattering and Rutherford's nuclear model, distance of closest approach and impact parameter, energy of an electron in a circular orbit, atomic spectra and the hydrogen series, Bohr's postulates, radius, speed and energy in the n-th orbit, the energy level diagram, de Broglie's explanation of the quantum condition, limitations of the Bohr model.

Alpha-particle scattering and the nuclear atom

In Thomson's model (1898) the positive charge of an atom was spread uniformly through it, with electrons embedded like seeds in a watermelon. In 1911, following Rutherford's suggestion, Geiger and Marsden fired 5.5 MeV alpha particles from a radioactive source at a thin gold foil (2.1 × 10−7 m thick) and counted the scattered particles on a zinc sulphide screen.

  • Most alpha particles went straight through or were deflected very slightly: about 0.14% were scattered by more than 1°.
  • About 1 in 8000 was deflected by more than 90°, and a few bounced almost straight back.

A spread-out positive charge could not produce such large deflections. Rutherford concluded that the entire positive charge and nearly all the mass of the atom are concentrated in a tiny nucleus, about 10−15 to 10−14 m across, while the atom itself is about 10−10 m across. Most of the atom is empty space. Electrons revolve round the nucleus, held by the Coulomb attraction, as planets revolve round the sun.

Impact parameter and closest approach

The impact parameter b is the perpendicular distance of the initial velocity of the alpha particle from the centre of the nucleus. A small b means a close encounter and a large deflection; b = 0 gives a head-on collision and the particle bounces straight back. For a head-on approach, the alpha particle stops momentarily at the distance of closest approach d, where all its kinetic energy K has become electrostatic potential energy:

K = 14πε0 · (2e)(Ze)d   ⇒   d = 2Ze24πε0K

d gives an upper limit on the size of the nucleus.

Worked example: An alpha particle of kinetic energy 7.7 MeV approaches a gold nucleus (Z = 79) head-on. Find the distance of closest approach.
Solution: d = (9 × 109 × 2 × 79 × (1.6 × 10−19)2)/(7.7 × 106 × 1.6 × 10−19) = (9 × 109 × 158 × 1.6 × 10−19)/(7.7 × 106) ≈ 3.0 × 10−14 m, about 30 fm.

Electron orbits

For an electron moving in a circle of radius r round a hydrogen nucleus, the Coulomb force provides the centripetal force: mv2/r = e2/(4πε0r2). This gives

K = e28πε0r,   U = −e24πε0r,   E = K + U = −e28πε0r

The total energy is negative, which means the electron is bound. Note that K = −E and U = 2E. Rutherford's model has a serious flaw: classical electromagnetism says an accelerating (orbiting) electron should radiate energy, spiral inwards and collapse into the nucleus, emitting a continuous spectrum. Atoms are stable and emit line spectra, so something else is needed.

Atomic spectra

A gas excited by an electric discharge emits light of only certain wavelengths, a line emission spectrum that is characteristic of the element. White light passed through the same cool gas shows dark lines at exactly those wavelengths (the absorption spectrum). For hydrogen, the lines fit the formula

1λ = R(1nf2 − 1ni2)R = 1.097 × 107 m−1 (Rydberg constant); ni > nf
SeriesnfniRegion
Lyman12, 3, 4, …Ultraviolet
Balmer23, 4, 5, …Visible
Paschen34, 5, 6, …Infrared
Brackett45, 6, 7, …Infrared
Pfund56, 7, 8, …Infrared

The first Balmer line, Hα (n = 3 to 2), is red at 656.3 nm. The lines of each series crowd together towards a series limit (ni → ∞).

Bohr model of the hydrogen atom

Bohr (1913) combined Rutherford's model with quantum ideas in three postulates:

  • An electron can revolve in certain stable orbits (stationary states) without radiating energy.
  • These are the orbits for which the angular momentum is an integral multiple of h/2π: L = mvr = nh/2π, n = 1, 2, 3, …
  • An electron jumping from a higher orbit to a lower one emits a photon with energy equal to the difference: hν = Ei − Ef.

Combining the quantum condition with the force equation gives, for hydrogen-like atoms of atomic number Z:

rn = n2Za0,   a0 = h2ε0πme2 = 5.29 × 10−11 mvn ∝ Zn   (v1 ≈ 2.2 × 106 m s−1 for hydrogen)En = −13.6 Z2n2 eV

a0 is the Bohr radius. The lowest state (n = 1) is the ground state; higher states are excited states. The energy needed to remove the electron from the ground state of hydrogen, the ionisation energy, is 13.6 eV. The energy to lift it from n = 1 to n = 2 (the first excitation energy) is 13.6 − 3.4 = 10.2 eV.

Energy level diagram of the hydrogen atomwww.iitmedicoguide.comn = ∞0 eVn = 5−0.54 eVn = 4−0.85 eVn = 3−1.51 eVn = 2−3.40 eVn = 1−13.6 eVLyman(ultraviolet)Balmer(visible)Paschen(infrared)Energy levels of hydrogen, En = −13.6/n² eV (not to scale)Arrows show emission; Brackett (to n = 4) and Pfund (to n = 5) lie further in the infraredwww.iitmedicoguide.com
Energy levels of hydrogen with the first three emission series. Transitions ending on n = 1 give the ultraviolet Lyman lines, those ending on n = 2 the visible Balmer lines, and those ending on n = 3 the infrared Paschen lines.

The line spectrum from Bohr's model

Putting En into hν = Ei − Ef gives exactly the Rydberg formula, with R expressed in terms of m, e, h and ε0. The calculated value agrees with the measured one. In absorption, an atom in the ground state absorbs only those photons whose energy matches a transition upwards; that is why the absorption spectrum of cool hydrogen shows only the Lyman lines.

If an atom is excited to level n, the number of different spectral lines it can emit on returning to the ground state is n(n − 1)/2.

Worked example: Find the wavelength of the Hα line and the energy needed to excite a hydrogen atom from its ground state to n = 3.
Solution: 1/λ = R(1/4 − 1/9) = 1.097 × 107 × 5/36 ≈ 1.524 × 106 m−1, so λ ≈ 656 nm. Excitation energy = E3 − E1 = −1.51 − (−13.6) = 12.09 eV.

de Broglie's explanation

Why should angular momentum be quantised? If the electron is a wave, a stable orbit must fit a whole number of wavelengths round its circumference, forming a standing wave: 2πrn = nλ. With λ = h/mv this becomes mvrn = nh/2π, which is Bohr's second postulate.

Limitations of the Bohr model

It works only for hydrogen and hydrogen-like ions (He+, Li2+) with a single electron; it cannot handle the interaction between electrons in larger atoms. It also cannot explain the relative intensities of spectral lines. These problems were resolved later by quantum mechanics.

Common mistakes: (1) Forgetting the Z2 in En for He+ or Li2+. (2) Treating the electron's total energy as positive; it is negative for a bound electron, and K = −E. (3) Mixing up ni and nf in the Rydberg formula and getting a negative wavelength. (4) Taking the shortest wavelength of a series from its first line; the shortest comes from the series limit (ni → ∞) and the longest from the first line. (5) Using the Bohr model for multi-electron atoms.

JEE and NEET focus

  • Observations and conclusions of the alpha scattering experiment; distance of closest approach.
  • Radius, speed, energy and their dependence on n and Z in the Bohr model.
  • Wavelengths of lines in the Lyman, Balmer and Paschen series, including series limits.
  • Ionisation and excitation energies, and the number of spectral lines from level n.
  • Relations among kinetic, potential and total energy of the electron.

Practice questions

In the alpha scattering experiment, most alpha particles pass through the foil undeflected. This shows that:

  1. the nucleus is positively charged
  2. most of the atom is empty space
  3. electrons are very light
  4. the atom is neutral
Show answer
B. Only a tiny nucleus can cause large deflections.

The radius of the third Bohr orbit of hydrogen is about:

  1. 1.59 Å
  2. 2.12 Å
  3. 4.77 Å
  4. 0.53 Å
Show answer
C. r = n2a0 = 9 × 0.529 Å.

The energy of the electron in the n = 2 state of He+ is:

  1. −3.4 eV
  2. −13.6 eV
  3. −54.4 eV
  4. −6.8 eV
Show answer
B. E = −13.6 × 22/22 eV.

The spectral series of hydrogen that lies in the visible region is:

  1. Lyman
  2. Balmer
  3. Paschen
  4. Pfund
Show answer
B. Transitions ending on n = 2.

The shortest wavelength in the Lyman series of hydrogen is about:

  1. 91 nm
  2. 122 nm
  3. 365 nm
  4. 656 nm
Show answer
A. 1/λ = R(1 − 0) gives λ = 1/R ≈ 91.2 nm.

For an electron in a Bohr orbit, the ratio of kinetic energy to total energy is:

  1. 1
  2. −1
  3. 2
  4. −2
Show answer
B. K = −E.

Hydrogen atoms are excited to n = 4. The number of different lines possible in the emission spectrum is:

  1. 3
  2. 4
  3. 6
  4. 10
Show answer
C. n(n − 1)/2 = 4 × 3/2.

The angular momentum of the electron in the second Bohr orbit is:

  1. h/2π
  2. h/π
  3. 2h
  4. h/4π
Show answer
B. L = nh/2π = 2h/2π.
Call WhatsApp Apply
Chat with us on WhatsApp