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Atoms

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Chapter Notes

Overview
This chapter traces the historical development of atomic models, from Thomson's plum pudding model to Rutherford's nuclear model and finally Bohr's quantum model of the hydrogen atom. It explains how alpha-particle scattering experiments led to the discovery of the nucleus, and how Bohr's postulates successfully explained the stability of atoms and the discrete line spectra of hydrogen. The one big idea a student must walk away with is that atoms have a tiny, dense, positively charged nucleus surrounded by electrons in quantized orbits, and that energy is emitted or absorbed only in discrete packets (photons) when electrons jump between these orbits.
Key Concepts (16)
Thomson's model (plum pudding model)
According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon.
e.g. Just like raisins are scattered throughout a pudding, electrons are embedded in a sphere of positive charge.
Rutherford's nuclear model (planetary model)
According to this model, the entire positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus, with electrons revolving around the nucleus just as planets revolve around the sun.
e.g. Like the sun at the centre of the solar system with planets orbiting around it, the nucleus is at the centre with electrons orbiting around it.
Alpha-particle (α-particle)
Alpha-particles are nuclei of helium atoms, carrying two units (2e) of positive charge and having the mass of a helium atom.
e.g. In the Geiger-Marsden experiment, α-particles from a radioactive source (²¹⁴₈₃Bi) were directed at a thin gold foil.
Impact parameter (b)
The perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus.
e.g. A small impact parameter means the α-particle passes very close to the nucleus and suffers large scattering; a large impact parameter means it goes nearly undeviated.
Distance of closest approach (d)
The centre-to-centre distance between the α-particle and the nucleus when the α-particle comes momentarily to rest and reverses its direction.
e.g. For a 7.7 MeV α-particle and a gold nucleus (Z=79), d = 3.0 × 10⁻¹⁴ m = 30 fm.
Stationary states
According to Bohr's first postulate, each atom has certain definite stable states in which it can exist without emitting radiant energy, and each possible state has definite total energy.
e.g. The n=1 state (ground state) of hydrogen is a stationary state with energy -13.6 eV.
Bohr's quantisation condition (second postulate)
The electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of h/2π, i.e., L = nh/2π, where n = 1, 2, 3, ... is the principal quantum number.
e.g. For n=1, L = h/2π; for n=2, L = 2h/2π = h/π, and so on.
Bohr's third postulate (frequency condition)
An electron can make a transition from a higher energy state (Ei) to a lower energy state (Ef), emitting a photon of energy hν = Ei – Ef.
e.g. When an electron jumps from n=2 (E₂ = -3.40 eV) to n=1 (E₁ = -13.6 eV), the emitted photon has energy 10.2 eV.
Ground state
The lowest energy state of an atom, with the electron revolving in the orbit of smallest radius (Bohr radius, a₀). For hydrogen, E₁ = -13.6 eV.
e.g. At room temperature, most hydrogen atoms are in the ground state (n=1).
Excited state
A state of the atom where the electron has been raised to a higher energy level (n > 1) by absorbing energy.
e.g. When a hydrogen atom absorbs 10.2 eV, it goes from n=1 (ground state) to n=2 (first excited state).
Ionisation energy
The minimum energy required to free the electron from the ground state of the hydrogen atom. For hydrogen, it is 13.6 eV.
e.g. If a hydrogen atom in ground state absorbs 13.6 eV or more, the electron is completely removed (ionised).
Emission line spectrum
A spectrum consisting of bright lines on a dark background, produced when excited atoms emit radiation at specific wavelengths as electrons fall from higher to lower energy states.
e.g. The Balmer series of hydrogen (visible lines) is an emission line spectrum.
Absorption spectrum
When white light passes through a gas, dark lines appear in the continuous spectrum at wavelengths corresponding to those absorbed by the atoms of the gas.
e.g. If white light passes through hydrogen gas, dark lines appear at the same wavelengths as the emission lines of hydrogen.
De Broglie wavelength (λ)
The wavelength associated with a material particle, given by λ = h/p, where p is the momentum of the particle.
e.g. For an electron moving at speed 2.2 × 10⁶ m/s in a hydrogen atom, λ = h/(mv).
Bohr radius (a₀)
The radius of the innermost orbit (n=1) of the hydrogen atom, equal to 5.3 × 10⁻¹¹ m.
e.g. The radius of the n=2 orbit is 4a₀ = 2.12 × 10⁻¹⁰ m, and of n=3 is 9a₀ = 4.77 × 10⁻¹⁰ m.
Hydrogenic atoms
Atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z is the proton number. Examples: hydrogen atom, singly ionised helium (He⁺), doubly ionised lithium (Li²⁺).
e.g. Bohr's model is applicable only to hydrogenic atoms, not to multi-electron atoms like helium.
Formulas & Equations (10)
F = (1/4πε₀) × (2e)(Ze)/r²
Reads as: F equals one over four pi epsilon naught times two e times Z e over r squared
Variables: F = electrostatic force of repulsion (N); ε₀ = permittivity of free space (8.85 × 10⁻¹² C²/Nm²); e = elementary charge (1.6 × 10⁻¹⁹ C); Z = atomic number of target nucleus; r = distance between α-particle and nucleus (m)
Use when: Calculating the force between an α-particle and a nucleus in Rutherford scattering
⚠ Forgetting that the α-particle has charge +2e, not +e, so the product is 2Ze², not Ze².
d = (1/4πε₀) × (2Ze²)/K
Reads as: d equals one over four pi epsilon naught times two Z e squared over K
Variables: d = distance of closest approach (m); ε₀ = permittivity of free space; Z = atomic number; e = elementary charge (C); K = initial kinetic energy of α-particle (J)
Use when: Finding the distance of closest approach of an α-particle to a nucleus
⚠ Using K in eV instead of joules; always convert eV to J (1 eV = 1.6 × 10⁻¹⁹ J).
mv²/r = (1/4πε₀) × e²/r²
Reads as: m v squared over r equals one over four pi epsilon naught times e squared over r squared
Variables: m = mass of electron (9.1 × 10⁻³¹ kg); v = velocity of electron (m/s); r = radius of orbit (m); ε₀ = permittivity of free space; e = elementary charge (C)
Use when: Relating centripetal force to electrostatic attraction for an electron in a hydrogen atom
⚠ Forgetting that the force is centripetal (mv²/r) and not centrifugal.
r = (4πε₀ × n²h²)/(4π²me²)
Reads as: r equals four pi epsilon naught times n squared h squared over four pi squared m e squared
Variables: r = radius of nth orbit (m); ε₀ = permittivity of free space; n = principal quantum number (1,2,3,...); h = Planck's constant (6.6 × 10⁻³⁴ Js); m = mass of electron (kg); e = elementary charge (C)
Use when: Calculating the radius of the nth Bohr orbit in hydrogen atom
⚠ Writing the formula as r = n²h²ε₀/(πme²) — the factor 4 is often missed. The correct simplified form is r = n² × (0.529 × 10⁻¹⁰ m).
Eₙ = -me⁴/(8ε₀²n²h²)
Reads as: E sub n equals negative m e to the fourth over eight epsilon naught squared n squared h squared
Variables: Eₙ = energy of nth stationary state (J); m = mass of electron (kg); e = elementary charge (C); ε₀ = permittivity of free space; n = principal quantum number; h = Planck's constant (Js)
Use when: Calculating the total energy of an electron in the nth orbit of hydrogen atom
⚠ Forgetting the negative sign — the negative sign indicates that the electron is bound to the nucleus.
Eₙ = -13.6/n² eV
Reads as: E sub n equals negative thirteen point six over n squared electron volts
Variables: Eₙ = energy of nth stationary state (eV); n = principal quantum number
Use when: Quickly finding the energy of any level in hydrogen atom without plugging in constants
⚠ Using this formula for multi-electron atoms — it is valid only for hydrogenic (single-electron) atoms.
hν = Eᵢ – E_f
Reads as: h nu equals E sub i minus E sub f
Variables: h = Planck's constant (6.6 × 10⁻³⁴ Js); ν = frequency of emitted photon (Hz); Eᵢ = energy of initial state (J or eV); E_f = energy of final state (J or eV)
Use when: Finding the frequency of radiation emitted when an electron jumps from a higher to a lower energy level
⚠ Writing hν = E_f – Eᵢ — the photon energy is the energy lost, so it is Eᵢ – E_f (since Eᵢ > E_f).
L = nh/2π
Reads as: L equals n h over two pi
Variables: L = angular momentum of electron (kg m²/s); n = principal quantum number (1,2,3,...); h = Planck's constant (Js)
Use when: Applying Bohr's quantisation condition for angular momentum
⚠ Writing L = nh — the correct expression is L = nh/2π.
2πrₙ = nλ
Reads as: two pi r sub n equals n lambda
Variables: rₙ = radius of nth orbit (m); n = integer (1,2,3,...); λ = de Broglie wavelength of electron (m)
Use when: Explaining de Broglie's standing wave condition for Bohr orbits
⚠ Forgetting that this is the condition for a standing wave — the circumference must be an integral multiple of the wavelength.
λ = h/p = h/(mv)
Reads as: lambda equals h over p equals h over m v
Variables: λ = de Broglie wavelength (m); h = Planck's constant (Js); p = momentum of particle (kg m/s); m = mass (kg); v = velocity (m/s)
Use when: Calculating the de Broglie wavelength of an electron or any material particle
⚠ Using the formula for photons (λ = hc/E) instead — for particles, use λ = h/p.
How It Works
Geiger-Marsden (Rutherford) alpha-particle scattering experiment
  1. Step 1: A beam of 5.5 MeV α-particles from a ²¹⁴₈₃Bi radioactive source is collimated into a narrow beam by passage through lead bricks.
  2. Step 2: The beam is allowed to fall on a thin gold foil of thickness 2.1 × 10⁻⁷ m.
  3. Step 3: Scattered α-particles are observed through a rotatable detector consisting of a zinc sulphide screen and a microscope.
  4. Step 4: Scattered α-particles on striking the screen produce brief light flashes (scintillations), which are viewed through the microscope.
  5. Step 5: The distribution of the number of scattered particles is studied as a function of the angle of scattering.
💡 Most α-particles pass through undeflected (about 99.86% scatter by less than 1°), only about 1 in 8000 deflect by more than 90°. This shows that most of the atom is empty space, and the positive charge is concentrated in a tiny nucleus.
Rutherford scattering — trajectory dependence on impact parameter
  1. Step 1: An α-particle approaches a gold nucleus with a given impact parameter b (perpendicular distance from initial velocity vector to nucleus centre).
  2. Step 2: The α-particle experiences a repulsive Coulomb force F = (1/4πε₀)(2Ze²/r²) directed along the line joining it to the nucleus.
  3. Step 3: For a small impact parameter (head-on collision), the α-particle comes very close to the nucleus, experiences a large repulsive force, and is scattered through a large angle (θ ≈ π, i.e., rebounds back).
  4. Step 4: For a large impact parameter, the α-particle passes far from the nucleus, experiences a weak force, and is scattered through a small angle (θ ≈ 0, nearly undeviated).
💡 The fact that only a small fraction of α-particles rebound back (small impact parameter collisions) implies that the mass and positive charge of the atom are concentrated in a very small volume — the nucleus.
Calculating distance of closest approach (d) using energy conservation
  1. Step 1: Write conservation of mechanical energy: initial kinetic energy K of α-particle = final electric potential energy U when α-particle momentarily stops.
  2. Step 2: U = (1/4πε₀) × (2e)(Ze)/d = (1/4πε₀) × (2Ze²)/d.
  3. Step 3: Set K = (1/4πε₀) × (2Ze²)/d.
  4. Step 4: Solve for d: d = (1/4πε₀) × (2Ze²)/K.
💡 The distance of closest approach gives an upper limit to the size of the nucleus — the actual nuclear radius is smaller than d because the α-particle reverses direction without touching the nucleus.
Bohr's model — deriving radius and energy of stationary orbits
  1. Step 1: Equate centripetal force to electrostatic attraction: mv²/r = (1/4πε₀)(e²/r²).
  2. Step 2: Apply Bohr's quantisation condition: mvr = nh/2π.
  3. Step 3: Solve the two equations simultaneously to eliminate v, obtaining rₙ = (4πε₀ n²h²)/(4π²me²).
  4. Step 4: Substitute rₙ into the expression for total energy E = K + U = (1/2)mv² - (1/4πε₀)(e²/r) to get Eₙ = -me⁴/(8ε₀²n²h²).
  5. Step 5: Substitute constants to get Eₙ = -13.6/n² eV.
💡 The negative total energy means the electron is bound to the nucleus. The ground state (n=1) has the most negative energy (-13.6 eV), meaning it is the most tightly bound.
Emission of a photon during electron transition
  1. Step 1: An electron in a higher energy state (Eᵢ, with quantum number nᵢ) spontaneously jumps to a lower energy state (E_f, with quantum number n_f).
  2. Step 2: The energy difference ΔE = Eᵢ – E_f is carried away by a single photon.
  3. Step 3: The frequency of the emitted photon is given by ν = ΔE/h = (Eᵢ – E_f)/h.
  4. Step 4: The wavelength of the emitted photon is λ = c/ν = hc/(Eᵢ – E_f).
💡 The photon is emitted only when the electron goes from a higher to a lower energy level. For absorption, the atom absorbs a photon of exactly the same energy to raise the electron to a higher level.
De Broglie's explanation of Bohr's quantisation condition
  1. Step 1: Assume the electron in its circular orbit behaves as a standing wave.
  2. Step 2: For a standing wave to persist, the circumference of the orbit must be an integral multiple of the de Broglie wavelength: 2πrₙ = nλ.
  3. Step 3: Substitute λ = h/(mvₙ) (de Broglie relation) into the standing wave condition: 2πrₙ = n × h/(mvₙ).
  4. Step 4: Rearrange to get mvₙrₙ = nh/2π, which is exactly Bohr's quantisation condition for angular momentum.
💡 Only those orbits where the electron wave forms a standing wave (constructive interference) are allowed — this is why angular momentum is quantised.
Exam Traps — Don't Make These Mistakes
⚠ Confusing distance of closest approach with nuclear radius
✗ The distance of closest approach d is equal to the radius of the nucleus.
→ The distance of closest approach is the centre-to-centre distance between the α-particle and the nucleus when the α-particle stops. The α-particle reverses direction without actually touching the nucleus, so d is greater than the sum of the radii of the α-particle and the nucleus. It gives only an upper limit to the nuclear size.
✓ The distance of closest approach gives an upper limit to the size of the nucleus. The actual nuclear radius is smaller than d.
⚠ Sign error in energy transition formula
✗ hν = E_f – Eᵢ
→ When an electron falls from a higher energy state (Eᵢ) to a lower energy state (E_f), the energy is lost by the atom, so the photon energy is Eᵢ – E_f (which is positive because Eᵢ > E_f). Writing E_f – Eᵢ gives a negative value, which is physically meaningless for emitted photon energy.
✓ hν = Eᵢ – E_f, where Eᵢ is the energy of the initial (higher) state and E_f is the energy of the final (lower) state.
⚠ Using Eₙ = -13.6/n² eV for multi-electron atoms
✗ The energy of the electron in helium atom in n=1 state is -13.6 eV.
→ The formula Eₙ = -13.6/n² eV is derived specifically for hydrogen (Z=1). For hydrogenic atoms with atomic number Z, the formula is Eₙ = -13.6 Z²/n² eV. For helium (Z=2), the ground state energy would be -54.4 eV, not -13.6 eV. Moreover, Bohr's model cannot be extended to multi-electron atoms like neutral helium.
✓ The formula Eₙ = -13.6/n² eV applies only to hydrogen atom. For hydrogenic atoms with atomic number Z, use Eₙ = -13.6 Z²/n² eV.
⚠ Confusing Thomson's and Rutherford's models regarding stability
✗ Rutherford's model is stable because electrons revolve around the nucleus like planets around the sun.
→ Rutherford's model is actually unstable according to classical electromagnetic theory. An accelerating charged particle (electron in circular orbit) must emit radiation, lose energy, and spiral into the nucleus. The planetary analogy is misleading because gravitational radiation is negligible, but electromagnetic radiation from accelerating charges is not.
✓ Rutherford's nuclear model is unstable because according to classical electromagnetic theory, an accelerating electron must continuously emit radiation, lose energy, and eventually fall into the nucleus. This contradicts the observed stability of atoms.
⚠ Forgetting that α-particle has charge +2e
✗ The force between α-particle and nucleus is F = (1/4πε₀)(e × Ze)/r² = (1/4πε₀)(Ze²)/r².
→ An α-particle is a helium nucleus with charge +2e, not +e. The correct force is F = (1/4πε₀)(2e × Ze)/r² = (1/4πε₀)(2Ze²)/r². Using e instead of 2e gives half the correct force.
✓ The α-particle carries two units of positive charge (2e), so the electrostatic force of repulsion is F = (1/4πε₀)(2Ze²)/r².
⚠ Mixing up emission and absorption spectra
✗ In an absorption spectrum, bright lines appear on a dark background.
→ An emission spectrum has bright lines on a dark background. An absorption spectrum has dark lines on a bright continuous background. The dark lines correspond to wavelengths that have been absorbed by the gas.
✓ An emission line spectrum consists of bright lines on a dark background. An absorption spectrum consists of dark lines on a continuous bright background.
Compare & Contrast
Distribution of positive charge and mass
Thomson's model (plum pudding model)
Rutherford's nuclear model
In Thomson's model, positive charge is uniformly distributed throughout the atom. In Rutherford's model, the entire positive charge and most of the mass are concentrated in a tiny nucleus at the centre.
How the spectrum is produced
Emission line spectrum
Absorption spectrum
An emission spectrum is produced when excited atoms emit photons as electrons fall from higher to lower energy levels, showing bright lines on a dark background. An absorption spectrum is produced when white light passes through a gas and atoms absorb photons of specific energies to raise electrons to higher levels, showing dark lines on a continuous bright background.
Stability of atom
Bohr's model
Classical (Rutherford) model
In the classical model, an accelerating electron must emit radiation and spiral into the nucleus, making the atom unstable. In Bohr's model, electrons can revolve in certain stationary orbits without emitting radiation, thus explaining atomic stability.
Exam Focus
Marks pattern: Typically 1 MCQ (1 mark) on atomic models or Bohr's postulates, 1 SA-I (2 marks) on distance of closest approach or energy calculation, 1 SA-II (3 marks) on derivation of radius or energy, and 1 LA (5 marks) on Bohr's model with numerical or derivation. Total: approximately 7-8 marks.
Most asked: Bohr's postulates and derivation of radius and energy of hydrogen atom · Rutherford's alpha-particle scattering experiment and conclusions · Distance of closest approach and impact parameter · Hydrogen line spectra (Balmer series, Lyman series, etc.) · De Broglie's explanation of Bohr's quantisation condition · Energy level diagram and transitions in hydrogen atom
Don't skip: Derive the expression for radius of nth Bohr orbit in hydrogen atom.; Derive the expression for total energy of electron in nth orbit of hydrogen atom.; Explain Rutherford's alpha-particle scattering experiment and its conclusions.; Calculate the wavelength or frequency of radiation emitted during a transition between two energy levels.; State and explain Bohr's postulates and de Broglie's explanation of the quantisation condition.
One-Line Revision (read aloud before the exam)
  1. Thomson's plum pudding model proposed a uniform positive sphere with embedded electrons.
  2. Rutherford's alpha-particle scattering experiment showed that most of the atom is empty space with a tiny, dense, positively charged nucleus.
  3. The impact parameter b is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus.
  4. The distance of closest approach d = (1/4πε₀)(2Ze²)/K gives an upper limit to nuclear size.
  5. Bohr's first postulate: electrons revolve in stationary orbits without emitting radiation.
  6. Bohr's second postulate: angular momentum is quantised as L = nh/2π.
  7. Bohr's third postulate: hν = Eᵢ – E_f when an electron jumps from a higher to a lower energy level.
  8. The radius of the nth Bohr orbit in hydrogen is rₙ = n² × 5.3 × 10⁻¹¹ m.
  9. The energy of the nth state in hydrogen is Eₙ = -13.6/n² eV.
  10. The ground state energy of hydrogen is -13.6 eV, and the ionisation energy is 13.6 eV.
  11. De Broglie explained Bohr's quantisation by requiring the electron orbit circumference to be an integral multiple of the de Broglie wavelength: 2πrₙ = nλ.
  12. Bohr's model is applicable only to hydrogenic (single-electron) atoms and cannot explain multi-electron atoms or relative intensities of spectral lines.
  13. An emission line spectrum has bright lines on a dark background; an absorption spectrum has dark lines on a continuous bright background.
  14. The Balmer series of hydrogen corresponds to transitions where the final state is n=2.
  15. The Lyman series of hydrogen lies in the ultraviolet region and corresponds to transitions to n=1.

MCQs — 10 Questions with Answers

Q1. In the Geiger-Marsden experiment, what was the primary target material used?

  • A. Silver foil
  • B. Gold foil of thickness 2.1 × 10⁻⁷ m ✓
  • C. Copper foil
  • D. Aluminium foil

Answer: B — The study material explicitly states a thin gold foil of thickness 2.1 × 10⁻⁷ m was used in the experiment.

Q2. Which observation from the alpha-scattering experiment most directly supported the existence of a concentrated nucleus?

  • A. Most α-particles passed through undeflected
  • B. Approximately 1 in 8000 α-particles deflected by more than 90° ✓
  • C. The beam intensity decreased uniformly across all angles
  • D. Alpha particles absorbed completely by the foil

Answer: B — Large backward deflection (>90°) can only result from close encounter with concentrated positive charge, proving nucleus concentration.

Q3. J.J. Thomson's plum pudding model proposed that electrons are:

  • A. Orbiting at discrete distances from nucleus
  • B. Embedded in uniformly distributed positive charge like seeds in watermelon ✓
  • C. Located only on the surface of atom
  • D. Concentrated at the centre with nucleus around them

Answer: B — Study material explicitly describes Thomson's 1898 plum pudding model as electrons embedded in uniformly distributed positive charge.

Q4. The atomic nucleus has a size of approximately 10⁻¹⁵ m. If the atom size is 10⁻¹⁰ m, by what factor is the atom larger than the nucleus?

  • A. 10² times
  • B. 10³ times
  • C. 10⁵ times ✓
  • D. 10⁷ times

Answer: C — Size ratio = 10⁻¹⁰ m ÷ 10⁻¹⁵ m = 10⁵; study material confirms atom is 10,000–100,000 times larger than nucleus.

Q5. What percentage of incident α-particles in Rutherford's experiment scattered by more than 1°?

  • A. 0.014%
  • B. 0.14% ✓
  • C. 1.4%
  • D. 14%

Answer: B — Study material explicitly states about 0.14% of incident α-particles scatter by more than 1° in the experiment.

Q6. Which statement is NOT correct regarding line spectra from rarefied gases?

  • A. Each element shows characteristic discrete wavelengths
  • B. Spectra consist of bright lines on dark background
  • C. Radiation is due to individual atoms with large average spacing
  • D. Continuous distribution of all wavelengths is observed ✓

Answer: D — Rarefied gases emit discrete line spectra, not continuous spectra; continuous spectra come from dense condensed matter.

Q7. A critical problem with Rutherford's nuclear model is that it cannot explain:

  • A. The size of the nucleus
  • B. The positive charge in atom
  • C. Why atoms emit discrete line spectra instead of continuous radiation ✓
  • D. The scattering of α-particles

Answer: C — Study material states Rutherford's model 'could not explain why atoms emit light of only discrete wavelengths' and could not explain stability of orbiting electrons radiating energy.

Q8. According to the study material, why does condensed matter emit continuous spectrum while rarefied gases emit line spectra?

  • A. Condensed matter has higher temperature
  • B. Condensed matter radiation is due to atomic oscillations from atom-atom interactions; rarefied gas radiation from individual atoms ✓
  • C. Rarefied gases have lower density and thus no interaction
  • D. Condensed matter absorbs only certain wavelengths

Answer: B — Study material explicitly explains continuous spectrum from condensed matter comes from oscillations governed by neighbor interactions, while line spectra from rarefied gases reflect individual atom emission.

Q9. In Rutherford's nuclear model, electrons are described as moving in orbits around nucleus similar to planets around sun. Which classical physics principle predicts this model should fail?

  • A. Newtons first law of motion
  • B. Conservation of energy — orbiting charged electron radiates EM energy and should spiral into nucleus ✓
  • C. Coulomb's law of electrostatics
  • D. Conservation of angular momentum only

Answer: B — Study material notes serious difficulty: orbiting electron should radiate EM energy per classical theory, lose energy, and spiral into nucleus, contradicting observed atomic stability.

Q10. The Balmer formula (1885) empirically described hydrogen line wavelengths but was significant because:

  • A. It proved atoms must contain electrons
  • B. It calculated nucleus size accurately
  • C. It suggested intimate relationship between atomic internal structure and discrete wavelengths, hinting at quantization not yet explained by Rutherford model ✓
  • D. It showed hydrogen spectrum was continuous

Answer: C — Study material states Balmer's formula 'suggested an intimate relationship between internal structure of atom and spectrum of radiation emitted' — a key hint for quantized energy levels.

Flashcards

What did J.J. Thomson's plum pudding model propose?

Positive charge uniformly distributed throughout atom volume with electrons embedded like seeds in watermelon.

Name the experiment that led to discovery of the atomic nucleus.

Geiger-Marsden alpha-particle scattering experiment (1911) on thin gold foil directed by Rutherford.

What fraction of α-particles scattered by more than 90° in Rutherford's experiment?

Approximately 1 in 8000 α-particles deflected by more than 90°.

What size range did Rutherford estimate for the atomic nucleus?

Nucleus size approximately 10⁻¹⁵ m to 10⁻¹⁴ m.

How many times larger is an atom compared to its nucleus?

Atom is about 10,000 to 100,000 times larger than nucleus diameter.

What does the characteristic spectrum of an element indicate?

Intimate relationship between internal atomic structure and discrete wavelengths of emitted radiation.

Why did most α-particles pass through the gold foil undeflected?

Atom is mostly empty space; only rare direct collisions with concentrated nucleus occur.

What observation from scattering suggested large repulsive force near nucleus?

Large backward deflection (>90°) of α-particles indicates close encounter with positive charge concentrated at centre.

Name the empirical formula relating hydrogen line wavelengths discovered in 1885.

Balmer formula obtained by Johann Jakob Balmer for group of hydrogen emission lines.

What is the fundamental difference between spectra of dense gases and rarefied gases?

Dense gases emit continuous spectrum (atomic interactions); rarefied gases emit discrete line spectrum (individual atoms).

Important Board Questions

State J.J. Thomson's plum pudding model of the atom. How did Rutherford's experiment disprove this model? [2 marks]

Define Thomson's model with uniformly distributed positive charge and embedded electrons. Then state that Rutherford's α-scattering showed large backward deflections impossible with uniform charge — nucleus must be concentrated and small.

Describe the Geiger-Marsden α-particle scattering experiment. What key observations led Rutherford to propose the nuclear model of the atom? [5 marks]

Describe apparatus: collimated α-beam on thin gold foil, detected scattered particles via scintillations. Key observations: (1) most pass undeflected — atom mostly empty space, (2) ~0.14% scatter >1°, (3) ~1 in 8000 deflect >90° — large repulsive force implies concentrated positive charge at centre. Large backward deflection only possible if positive charge tightly concentrated; this led to nuclear model.

Explain why Rutherford's nuclear model, despite successfully explaining α-scattering, could not account for the discrete line spectra observed in rarefied gases. What fundamental classical physics principle is violated? [6 marks]

Rutherford's model has orbiting electron around nucleus like planet around sun. According to classical electromagnetism, orbiting charged particle radiates EM energy continuously, loses energy, spirals into nucleus — predicts atomic collapse and continuous emission spectrum. But atoms are stable and emit discrete lines only. This contradiction (no mechanism for quantization) revealed need for quantum theory. Show how classical acceleration of orbiting electron predicts continuous radiation, contradicting observed discrete spectra.

Assertion-Reason Questions for Atoms

Q1. Assertion (A):

Rutherford's nuclear model successfully explained why α-particles could pass through the thin gold foil almost without deflection.

Reason (R):

The nucleus occupies only a tiny fraction of the atom's volume (about 10⁻¹⁵ of total volume), so the probability of an α-particle approaching the nucleus closely is extremely small.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why most α-particles pass through undeflected. The nuclear model predicts that the atom is mostly empty space, making direct nuclear encounters rare. This is the fundamental insight that reconciles the observation with the nuclear structure hypothesis.

Q2. Assertion (A):

Thomson's plum pudding model could have explained the results of the Geiger-Marsden scattering experiment.

Reason (R):

In the plum pudding model, the positive charge is uniformly distributed, so α-particles would experience similar deflecting forces regardless of their impact parameter.

Answer: D — A is false but R is true — The assertion is false. The plum pudding model cannot explain the observation that some α-particles are deflected at very large angles (90°+) while most pass through undeflected. The reason is true in describing the model, but since the assertion itself is wrong, the overall answer is D. The experimental data disproved this model conclusively.

Q3. Assertion (A):

The observation that about 1 in 8000 α-particles deflected by more than 90° is consistent with a nucleus of size 10⁻¹⁵ m in an atom of size 10⁻¹⁰ m.

Reason (R):

The small fraction of large-angle scattering is expected because α-particles can only be strongly deflected if they pass very close to the tiny nucleus.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true and correctly related. The size ratio of ~10⁵ means nuclear encounters are rare (proportional to nuclear cross-section), predicting that large deflections occur only for a tiny fraction of incident particles. This quantitative consistency between size estimates and scattering probability validates the nuclear model.

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Case-Based Questions (CBQ) for Atoms

Case Study:

A research team in Mumbai conducted a modern version of the Geiger-Marsden experiment using 5.5 MeV α-particles from a ²¹⁴Bi radioactive source, directed at a thin silver foil (thickness 1.5 × 10⁻⁷ m). They detected scattered α-particles at various angles using a scintillation counter. The scattering rate at different angles (θ) was recorded: at θ = 30°, 450 particles/minute; at θ = 60°, 120 particles/minute; at θ = 90°, 25 particles/minute; at θ = 120°, 8 particles/minute. The theoretical prediction based on Rutherford's formula for 1/sin⁴(θ/2) matched the experimental data within 5% error.

Q1. What does the rapid decrease in scattered particle count as scattering angle increases tell us about the nuclear model?

  • A. The nucleus repels α-particles very weakly at all distances
  • B. Large-angle scattering is rare because the nucleus is very small, and close encounters are uncommon ✓
  • C. Most α-particles are absorbed by the silver foil
  • D. The nuclear charge decreases with scattering angle

Answer: B

Q2. The data shows 450 particles at 30° but only 8 at 120°. What is the approximate ratio of scattering rates (Rate₃₀°/Rate₁₂₀°)?

  • A. 56.25 (approximately) ✓
  • B. 15 (approximately)
  • C. 450 (approximately)
  • D. 3.2 (approximately)

Answer: A

Q3. Rutherford's scattering formula predicts that the number of particles scattered is proportional to 1/sin⁴(θ/2). Verify this for the given data at θ = 30° and θ = 90°. (sin 15° ≈ 0.259, sin 45° ≈ 0.707)

  • A. The ratio of rates matches 1/sin⁴(15°) ÷ 1/sin⁴(45°) ≈ 56, confirming Rutherford's prediction ✓
  • B. The ratio does not match because the foil is too thick
  • C. The prediction works only for forward scattering, not backward
  • D. Silver foil violates Rutherford's assumptions

Answer: A

Q4. If this experiment had used a gold foil (Z = 79) instead of silver (Z = 47), how would the scattering rates at the same angles compare?

  • A. Scattering rates would remain unchanged because the nuclear charge does not affect Coulomb scattering
  • B. Scattering rates would increase (approximately in proportion to Z²) because the higher nuclear charge produces stronger Coulomb repulsion ✓
  • C. Scattering rates would decrease because heavier nuclei are more stable
  • D. Only the maximum scattering angle would increase

Answer: B

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Previous Year Questions (PYQ) for Atoms

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

Assertion (A): In Bohr model of hydrogen atom, the energy levels are discrete and quantised. Reason (R): In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Both Assertion (A) and Reason (R) are false.

CBSE 2026 · Set 55/1/3 · 1 mark

The ratio of the potential energy to the kinetic energy of an electron in nᵗʰ orbit of Bohr model of hydrogen atom is

  • −1/2
  • 1/2
  • 2
  • −2

CBSE 2026 · Set 55/2/1 · 1 mark

In Bohr model of hydrogen atom, for large values of n, the distance between the consecutive orbits is proportional to

  • √n
  • n
  • n²
  • n³

CBSE 2026 · Set 55/2/1 · 1 mark

In Bohr model of hydrogen atom, the electron makes a transition from n = 5 to n = 1 state. As a result, a photon of wavelength λ is emitted in the process. The wavelength of the photon emitted when an electron makes a transition from energy level n = 5 to n = 2 will be

  • 8λ/7
  • 24λ/7
  • 16λ/7
  • 32λ/7

CBSE 2026 · Set 55/2/1 · 3 marks

(a) State Bohr's second postulate and mention its significance. (b) Prove that, in Bohr model of hydrogen atom as principal quantum number n becomes large, the energy levels get closer and closer.

CBSE 2026 · Set 55/2/2 · 1 mark

In Bohr model of hydrogen atom, the value of potential energy of an electron in nᵗʰ orbit varies with 'n' as

  • 1/n²
  • 1/n
  • n
  • n²

CBSE 2026 · Set 55/2/2 · 1 mark

In Bohr model of hydrogen atom, an electron makes a transition from n = 4 state to n = 1 state and a photon of frequency ν is emitted. The frequency of photon emitted when an electron makes a transition from n = 4 state to n = 2 state in the same model is

  • ν/3
  • ν/4
  • ν/5
  • ν/6

CBSE 2026 · Set 55/2/2 · 3 marks

Calculate the de Broglie wavelength associated with an electron revolving in the second excited state of hydrogen atom. The ground state energy of the hydrogen atom is −13.6 eV. Take m_e = 9 × 10⁻³¹ kg, h = 6.6 × 10⁻³⁴ J.s.

CBSE 2026 · Set 55/2/3 · 1 mark

In Bohr model of hydrogen atom, an electron is moving in an orbit, with total energy −3.4 eV. Its angular momentum will be

  • 1.05 × 10⁻³⁴ J s
  • 6.63 × 10⁻³⁴ J s
  • 3.31 × 10⁻³⁴ J s
  • 2.1 × 10⁻³⁴ J s

CBSE 2026 · Set 55/2/3 · 1 mark

According to Bohr model of hydrogen atom, the time period of an electron in its nᵗʰ orbit is proportional to

  • n
  • n²
  • n³
  • n⁴

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

Suppose there is an atom that contains five energy levels. How many transitions are possible in which photons are emitted ?

  • 4
  • 5
  • 10
  • 15

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

In Bohr's model of hydrogen atom, the potential energy of the electron in the ground state is –

  • −27.2 eV
  • 27.2 eV
  • −13.6 eV
  • 13.6 eV

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Class 12 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Physics Previous Year Questions, Chapter-wise

All 962 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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Class 12 Physics Derivations

All 68 derivations, chapter-wise, step by step with diagrams and marked with the CBSE board years (2022–2026) each was asked in. Free PDF download.

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