NEET Physics · NCERT Class 12

Atoms notes

Rutherford Model, Atomic Spectra, Bohr Model, Hydrogen Spectrum and de Broglie Explanation — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 6 topics
  • 44 min read
  • 23 formulas
  • 26 PYQs
1

Chapter Overview

6 min

The chapter Atoms explains how our idea of atomic structure developed from Rutherford’s nuclear atom to Bohr’s quantized model and finally to the wave nature of electrons. Rutherford’s alpha scattering experiment proved that the atom has a tiny, dense, positively charged nucleus. Atomic spectra revealed that atoms emit or absorb only definite frequencies, which classical physics could not explain. Bohr introduced stationary orbits, quantized angular momentum, energy levels, excitation and ionization. The hydrogen spectrum is understood through transitions between energy levels using the Rydberg formula. de Broglie’s matter-wave idea explains why only certain electron orbits are allowed. For NEET, this chapter is formula-heavy, concept-based, and frequently tested through direct numerical questions.

Key points7

  • The atom is electrically neutral but contains a tiny positive nucleus and surrounding electrons.
  • Classical mechanics fails for atomic stability and line spectra.
  • Bohr combined Rutherford’s nuclear model with Planck’s quantum idea.
  • Only transitions between allowed energy levels produce spectral lines.
  • The Rydberg formula connects wavelength with initial and final quantum numbers.
  • Matter waves give a physical explanation for stable electron orbits as standing waves.
  • NEET often asks radius, velocity, energy, wavelength, frequency, excitation energy and ionization energy.

Memory tricks2

Chapter Logic Trick: Remember the order as R-S-B-H-D: Rutherford sees nucleus, Spectra show lines, Bohr explains orbits, Hydrogen gives series, de Broglie explains why orbits are allowed.

Energy Sign Trick: Bound electron energy is negative. Zero energy means the electron is just free from the atom.

Examples2

NEET-Style Direct Fact

The ground state energy of hydrogen is -13.6 eV. Energy needed to ionize it from ground state is +13.6 eV.

Real-Life Link

Different colours in discharge tubes and neon signs occur because atoms emit light of characteristic wavelengths.

Common mistakes3

Using Bohr Model for All Atoms

Bohr formulas are directly valid only for hydrogen and hydrogen-like one-electron species. Do not apply them to multi-electron atoms unless explicitly approximated.

Confusing Excitation and Ionization

Excitation means electron moves to a higher bound orbit. Ionization means electron is completely removed to n = ∞.

Wrong Transition Direction

Emission occurs when electron moves from higher n to lower n. Absorption occurs when it moves from lower n to higher n.

Reference tables2

Quick revision8

Formulas4

Rydberg Formula
1/λ=RZ2(1/n12−1/n22)1/λ = RZ²(1/n₁² - 1/n₂²)
n2>n1n₂ > n₁

Gives the wavelength of radiation emitted or absorbed during transition in a hydrogen-like atom.

λ
Wavelength of spectral line
R
Rydberg constant, approximately 1.097 × 10⁷ m⁻¹
Z
Atomic number of hydrogen-like species
n₁
Lower or final energy level
n₂
Higher or initial energy level
Bohr Radius
rn=0.529n2/ZA˚rₙ = 0.529 n²/Z Å

Radius of the nth permitted orbit for a hydrogen-like atom.

rₙ
Radius of nth Bohr orbit
n
Principal quantum number
Z
Atomic number
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Diagrams3

2

Rutherford Model

7 min

Rutherford’s model came from the alpha particle scattering experiment, where fast alpha particles were directed at a very thin gold foil. Most alpha particles passed straight through, a few were slightly deflected, and a very small number rebounded through large angles. Rutherford concluded that most of the atom is empty space, positive charge and nearly all mass are concentrated in a tiny central nucleus, and electrons revolve around it. The nuclear size was found to be much smaller than atomic size, about 10⁻¹⁵ m compared with atomic size around 10⁻¹⁰ m. However, the model failed to explain atomic stability and line spectra because revolving electrons should radiate energy continuously according to classical electromagnetic theory.

Key points6

  • The experiment used a radioactive alpha source, a thin gold foil and a fluorescent zinc sulphide screen.
  • Scattering is due to electrostatic repulsion between alpha particles and the positive nucleus.
  • The probability of large deflection is very small because the nucleus occupies very little volume.
  • The nuclear model replaced Thomson’s plum pudding model.
  • Classical physics predicts an accelerating electron should radiate energy, making Rutherford’s atom unstable.
  • Rutherford model provided the foundation for Bohr’s model.

Memory tricks2

Observation Trick: Most-Some-Few: Most passed straight, Some deflected, Few bounced back. This directly gives Empty atom, Positive centre, Tiny dense nucleus.

Size Trick: Atom is 10⁻¹⁰ m and nucleus is 10⁻¹⁵ m. Remember: atom has 10, nucleus has 15; nucleus is smaller by 10⁵ in radius.

Examples2

Analogy

If the atom were a huge stadium, the nucleus would be like a tiny object near the centre; most of the stadium would be empty.

NEET Numerical Sense

If atomic radius is 10⁻¹⁰ m and nuclear radius is 10⁻¹⁵ m, the ratio of radii is 10⁵ and the ratio of volumes is about 10¹⁵.

Common mistakes3

Thinking Alpha Particles Are Electrons

Alpha particles are positively charged helium nuclei, not electrons. Their positive charge causes repulsion from the positive nucleus.

Saying Nucleus Contains All Atomic Volume

The nucleus contains nearly all mass but occupies a tiny fraction of volume.

Ignoring Classical Instability

Rutherford model fails because revolving electrons are accelerated charges and should continuously lose energy.

Reference tables2

Quick revision8

Formulas2

Order of Nuclear Radius
R=R0A(1/3)R = R₀A^(1/3)

Approximate relation for nuclear radius, showing that nuclear size depends on mass number.

R
Radius of nucleus
R₀
Constant of order 1.2 × 10⁻¹⁵ m
A
Mass number
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Diagrams3

3

Atomic Spectra

7 min

Atomic spectra are the patterns of light obtained when radiation from atoms is separated by a prism or diffraction grating. A hot solid, liquid, or dense gas usually gives a continuous spectrum containing all wavelengths. A low-pressure excited gas gives an emission line spectrum consisting of bright lines at specific wavelengths. When white light passes through a cooler gas, the gas absorbs its characteristic wavelengths and forms a dark-line absorption spectrum. For isolated atoms, spectra are not continuous because atomic energy is quantized. Hydrogen shows several spectral series produced by electron transitions to fixed lower levels. These spectral lines were among the strongest evidences against classical atomic models and led to Bohr’s quantized energy levels.

Key points7

  • Spectra are formed by transitions between quantized atomic energy levels.
  • Emission happens when an electron falls from higher to lower energy.
  • Absorption happens when an electron absorbs a photon and jumps to higher energy.
  • The same element absorbs the same wavelengths that it emits.
  • Line spectra are used to identify elements in stars and gases.
  • A smaller wavelength corresponds to higher frequency and higher photon energy.
  • Hydrogen spectrum is mathematically described by the Rydberg formula.

Memory tricks2

Emission vs Absorption: Emission = Exit energy as light, electron comes down. Absorption = Accept energy, electron goes up.

Spectrum Identity: Think of line spectrum as an atomic barcode: every element has its own fixed pattern of lines.

Examples2

Astronomy Example

The elements present in stars are identified by comparing their absorption lines with laboratory spectra.

Discharge Tube Example

Hydrogen gas in a discharge tube emits characteristic coloured lines because electrons de-excite between fixed levels.

Common mistakes3

Calling Hydrogen Spectrum Continuous

Hydrogen gives a line spectrum, not a continuous spectrum, because electron energies are quantized.

Wrong Energy-Wavelength Relation

Higher energy means higher frequency but smaller wavelength. Do not say high energy means large wavelength.

Mixing Bright and Dark Lines

Emission lines are bright. Absorption lines are dark at the same wavelengths for the same gas.

Reference tables2

Quick revision8

Formulas3

Photon Energy
E=hν=hc/λE = hν = hc/λ

Energy carried by a photon of frequency ν or wavelength λ.

E
Photon energy
h
Planck constant
ν
Frequency of radiation
c
Speed of light
λ
Wavelength
Energy Difference in Transition
ΔE=E2−E1=hνΔE = E₂ - E₁ = hν

The photon energy equals the difference between two atomic energy levels.

ΔE
Difference between two energy levels
E₂
Higher energy level
E₁
Lower energy level
ν
Frequency of emitted or absorbed radiation
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Diagrams3

4

Bohr Model

9 min

Bohr’s model corrected Rutherford’s instability problem by introducing quantum postulates. According to Bohr, electrons revolve around the nucleus only in certain permitted stationary orbits without radiating energy. The angular momentum of the electron is quantized as mvr = nh/2π. Radiation is emitted or absorbed only when an electron jumps between two stationary orbits, with photon energy hν equal to the energy difference. For hydrogen-like atoms, Bohr derived formulas for orbit radius, electron velocity and energy. The model explains hydrogen spectrum, excitation energy and ionization energy very well. Its limitations are that it fails for multi-electron atoms, fine structure, Zeeman effect and the full wave nature of electrons.

Key points7

  • Bohr postulates combine Rutherford’s nucleus with Planck’s quantum theory.
  • Negative energy means the electron is bound to the nucleus.
  • As n increases, radius increases and total energy approaches zero.
  • Ground state is n = 1 and has minimum energy.
  • Excited states have n > 1.
  • Ionization corresponds to n = ∞ where E = 0.
  • For hydrogen, first excitation energy is 10.2 eV and ionization energy is 13.6 eV.

Memory tricks3

Bohr Postulates: Remember S-Q-J: Stationary orbits, Quantized angular momentum, Jump radiation.

Radius-Energy Trend: r grows with n², energy goes toward zero. Larger orbit means less tightly bound electron.

Hydrogen Energies: 13.6, 3.4, 1.51 eV are magnitudes for n = 1, 2, 3 because divide 13.6 by n².

Examples3

Solved Numerical: Radius

For He+ in n=2, r₂ = 0.529 × 2²/2 Å = 1.058 Å.

Solved Numerical: Energy

For Li2+ in n=3, E₃ = -13.6 × 3²/3² = -13.6 eV.

Solved Numerical: First Excitation of Hydrogen

E₁ = -13.6 eV and E₂ = -3.4 eV, so excitation energy = 10.2 eV.

Common mistakes3

Forgetting Negative Sign in Energy

Eₙ is negative for bound states. Ionization energy is positive because it is energy supplied.

Confusing n₁ and n₂

In emission, n₂ is initial higher level and n₁ is final lower level. Always ensure n₂ > n₁ in the Rydberg formula.

Applying Bohr to Multi-Electron Atoms

Bohr formulas are not exact for atoms like helium atom or lithium atom; they are exact only for one-electron ions such as He+ and Li2+.

Reference tables2

Quick revision8

Formulas6

Quantization of Angular Momentum
mvr=nh/2π=nℏmvr = nh/2π = nℏ

Only those orbits are allowed in which angular momentum is an integral multiple of ℏ.

m
Mass of electron
v
Speed of electron
r
Radius of orbit
n
Principal quantum number
ℏ
Reduced Planck constant, h/2π
Bohr Radius of nth Orbit
rn=0.529n2/ZA˚rₙ = 0.529 n²/Z Å

Radius of nth orbit in a hydrogen-like atom.

rₙ
Radius of nth orbit
n
Orbit number
Z
Atomic number
Electron Velocity in nth Orbit
vn=2.18×106Z/nms−1vₙ = 2.18 × 10⁶ Z/n m s⁻¹

Speed of electron in nth orbit of a hydrogen-like atom.

vₙ
Electron speed in nth orbit
Z
Atomic number
n
Orbit number
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Diagrams3

5

Hydrogen Spectrum

8 min

Hydrogen spectrum is the most important application of Bohr’s model. When an electron in hydrogen jumps from a higher energy level to a lower energy level, it emits a photon whose wavelength is determined by the Rydberg formula. A group of lines ending at the same lower level is called a spectral series. Transitions ending at n = 1 form the Lyman series in the ultraviolet region. Transitions ending at n = 2 form the Balmer series, partly visible. Transitions ending at n = 3, 4 and 5 form the Paschen, Brackett and Pfund series respectively in the infrared region. Series limit occurs when the initial level n₂ becomes infinity, giving the shortest wavelength and maximum energy in that series.

Key points7

  • Rydberg formula is the central formula for hydrogen spectrum.
  • For emission spectrum, electron falls from n₂ to n₁ where n₂ > n₁.
  • For a fixed n₁, as n₂ increases, wavelength decreases and approaches the series limit.
  • Lyman has the highest energy photons among hydrogen series because transitions end at n = 1.
  • Balmer is important because many of its lines are visible.
  • The first line of a series occurs for n₂ = n₁ + 1 and has the longest wavelength in that series.
  • The limiting line occurs for n₂ = ∞ and has the shortest wavelength.

Memory tricks3

Series Order: Remember L-B-P-B-P as Lazy Boys Play Better Piano: Lyman, Balmer, Paschen, Brackett, Pfund.

Final n Trick: Series final levels are simply 1, 2, 3, 4, 5 in the same order: Lyman 1, Balmer 2, Paschen 3, Brackett 4, Pfund 5.

Region Trick: Only Balmer is famous for visible lines. Lyman is UV; Paschen, Brackett and Pfund are IR.

Examples3

Solved Numerical: Lyman First Line

For transition 2→1 in hydrogen, 1/λ = R(1 - 1/4) = 3R/4.

Solved Numerical: Balmer Series Limit

For Balmer limit, n₁ = 2 and n₂ = ∞, so 1/λ = R/4 and λ = 4/R.

Identification Example

A transition ending at n = 3 belongs to the Paschen series, regardless of whether it starts from n = 4, 5, 6 or higher.

Common mistakes3

Confusing Initial and Final Levels

The series name depends on final level n₁, not the initial level n₂.

Wrong Series Limit

Series limit is not the first line. It occurs for n₂ = ∞ and gives the shortest wavelength.

Saying All Balmer Lines Are Visible

Balmer series is associated with the visible region, but not every Balmer transition must be treated as visibly coloured in all contexts.

Reference tables2

Quick revision8

Formulas3

Hydrogen Rydberg Formula
1/λ=R(1/n12−1/n22)1/λ = R(1/n₁² - 1/n₂²)

Wavelength of hydrogen spectral line for transition from n₂ to n₁.

λ
Wavelength of spectral line
R
Rydberg constant, 1.097 × 10⁷ m⁻¹
n₁
Final lower orbit
n₂
Initial higher orbit
Series Limit
1/λlimit=R/n121/λ_{\text{limit}} = R/n₁²

Obtained when n₂ = ∞ for a fixed final level n₁.

λ_limit
Shortest wavelength of the series
R
Rydberg constant
n₁
Final orbit defining the series
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6

de Broglie Explanation

7 min

de Broglie proposed that moving particles have wave nature along with particle nature. The wavelength associated with a particle is λ = h/p = h/mv. For an electron moving in a Bohr orbit, only those circular paths are stable in which the electron wave forms a standing wave. This means the circumference of the orbit must contain a whole number of wavelengths: 2πr = nλ. Substituting λ = h/mv gives mvr = nh/2π, which is exactly Bohr’s quantization of angular momentum. Thus de Broglie’s idea gives a physical explanation for why only certain electron orbits are allowed. Wave-particle duality is experimentally supported by electron diffraction and is fundamental to modern quantum physics.

Key points7

  • Wave nature is significant for microscopic particles like electrons because their wavelength is measurable.
  • For macroscopic objects, de Broglie wavelength is extremely small and not noticeable.
  • A stable electron orbit is possible only when the wave joins smoothly with itself.
  • If circumference is not an integral multiple of wavelength, destructive interference makes the orbit unstable.
  • de Broglie idea connects particle momentum with wave wavelength.
  • Wave-particle duality does not mean the electron is sometimes only wave and sometimes only particle; it shows both aspects depending on experiment.
  • This explanation supports Bohr’s postulate instead of assuming it without reason.

Memory tricks3

Formula Trick: de Broglie connects Wave and Particle: λ is wave, p is particle, and h is the bridge: λ = h/p.

Allowed Orbit Trick: Allowed orbit means the electron wave says, 'I must meet myself perfectly after one round.' That is 2πr = nλ.

Momentum-Wavelength Trick: Fast and heavy means high momentum, so wavelength becomes tiny.

Examples3

Solved Numerical: Electron Wavelength

If an electron has momentum p = 6.63 × 10⁻²⁴ kg m/s, then λ = h/p = 6.63 × 10⁻³⁴ / 6.63 × 10⁻²⁴ = 10⁻¹⁰ m.

Bohr Quantization Derivation

From 2πr = nλ and λ = h/mv, we get 2πr = nh/mv. Rearranging gives mvr = nh/2π.

Real-Life Experimental Significance

Electron diffraction by crystals confirms that electrons behave like waves, because diffraction is a wave phenomenon.

Common mistakes3

Using λ = h/mv for Photons

For photons, use p = h/λ and E = pc. The formula λ = h/mv is for material particles with rest mass in non-relativistic motion.

Ignoring Units

Use mass in kg, speed in m/s and h in J s to get wavelength in metre.

Thinking Any Circular Orbit Is Allowed

Only orbits where circumference equals an integral number of wavelengths are allowed.

Reference tables2

Quick revision8

Formulas5

de Broglie Wavelength
λ=h/pλ = h/p

Wavelength associated with any moving particle.

λ
Matter wave wavelength
h
Planck constant
p
Momentum of particle
Non-relativistic de Broglie Wavelength
λ=h/mvλ = h/mv

Used when a particle of mass m moves with speed v much less than speed of light.

m
Mass of particle
v
Speed of particle
h
Planck constant
λ
de Broglie wavelength
Electron Accelerated Through Potential Difference
λ=h/√(2meV)λ = h/√(2meV)

Wavelength of an electron accelerated from rest through potential difference V.

λ
de Broglie wavelength of electron
h
Planck constant
m
Mass of electron
e
Electronic charge
V
Accelerating potential difference
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Diagrams3

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NEET PYQs from Atoms

The latest 3 of 26 previous-year questions from this chapter.

  1. NEET 2026Set 50MediumQ1

    Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5\ \mathrm{ms^{-1}}$. The radius of the orbit is $x \times 10^{-9}\ \mathrm{m}$. The value of $x$ is: [Take the mass of electron to the $9 \times 10^{-31}\ \mathrm{kg}$, charge of electron = $-1.6 \times 10^{-19}\ \mathrm{C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \mathrm{Nm^2C^{-2}}$]

    • A$4$
    • B$3$
    • C$2$
    • D$1$
  2. NEET 2026Set 50EasyQ2

    In Geiger-Marsden experiment, the number of scattered $\alpha$-particles $N(\theta)$ is plotted as a function of scattering angle $\theta$. Which of the following options represents the correct plot?

    • AGraph A
    • BGraph B
    • CGraph C
    • DGraph D
  3. NEET 2025Set 45HardQ3

    A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius $r$ of the $n^{th}$ orbit and the particle’s speed $v$ in the orbit depend on $n$ as

    • A$r \propto n^{1/3},\; v \propto n^{1/3}$
    • B$r \propto n^{1/3},\; v \propto n^{2/3}$
    • C$r \propto n^{2/3},\; v \propto n^{1/3}$
    • D$r \propto n^{4/3},\; v \propto n^{-1/3}$

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