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NEET Physics - Atoms Complete Study Notes

NEET Physics: Atoms & Bohr Model Master Notes

Unit: Modern Physics | Weightage: 2 - 3 Questions (8 - 12 Marks) | Target Exam: NEET UG

1. Rutherford Alpha Particle Scattering Experiment

In 1911, Hans Geiger and Ernest Marsden, under the guidance of Ernest Rutherford, conducted the landmark alpha particle scattering experiment using a thin gold foil of thickness 2.1 × 10−7 m (about 400 atoms thick).

Key Experimental Observations

  • > 99.86% of α-particles passed straight through the foil with virtually zero deflection (θ < 1°).
  • Only ~ 1 in 8,000 α-particles suffered a deflection of more than 90°.
  • Very rare (~ 1 in 106) was deflected by nearly 180° (head-on collision rebound).
Rutherford's Inferences:
  1. Most of an atom is empty space (hollow).
  2. The entire positive charge and almost 99.9% of the mass of an atom is concentrated in a tiny core called the Nucleus.
  3. Nuclear radius (r0 ∼ 10−15 m = 1 fm) is roughly 10−5 times smaller than the atomic radius (Ra ∼ 10−10 m = 1 Å).
  4. Volume ratio of atom to nucleus: VatomVnucleus ∼ (10−10)3(10−15)3 = 1015.
Rutherford Alpha Particle Scattering Experiment, Distance of Closest Approach, and Impact Parameter
Figure 1: Geiger-Marsden Alpha Scattering Geometry, Trajectories, Closest Approach r0, and Impact Parameter b

Distance of Closest Approach (r0)

For an α-particle fired directly towards a gold nucleus (head-on collision, impact parameter b = 0), all its initial kinetic energy Ek is converted into electrostatic potential energy U at the turning point:

Distance of Closest Approach Formula
Ek = 12 m v2 = 14πε0 (2e)(Ze)r0
r0 = 14πε0 2 Z e2Ek = 4 k Z e2m v2
NEET High-Yield Trap: Note that r0 ∝ ZEk ∝ 1v2. If the velocity of the incident α-particle is doubled, the distance of closest approach becomes one-fourth (r0 / 4)!

Impact Parameter (b)

The impact parameter b is defined as the perpendicular distance of the initial velocity vector of the α-particle from the center of the target nucleus:

Impact Parameter & Scattering Angle Relation
b = 14πε0 Z e2 cot(θ2)Ek
  • When b = 0 (head-on collision): cot(θ / 2) = 0 ⇒ θ = 180° (rebound).
  • When b is very large: cot(θ / 2) → ∞ ⇒ θ → 0° (undeflected).

Rutherford Scattering Formula

The number of α-particles N(θ) scattered at an angle θ per unit area onto the detector is:

Rutherford Angular Scattering Distribution
N(θ) ∝ 1sin4(θ2)  ⇒  N(θ1)N(θ2) = sin4(θ22)sin4(θ12)

2. Rutherford's Model Limitations & The Need for Bohr's Model

Two Fatal Limitations of Rutherford's Model:
  1. Atomic Instability: An electron in a circular planetary orbit is constantly accelerated towards the center (a = v2 / r). According to classical Maxwellian electrodynamics, an accelerated charged particle must continuously radiate electromagnetic energy. Hence, the electron should lose energy, spiral inward, and crash into the nucleus in ~ 10−8 seconds. Real atoms are remarkably stable!
  2. Line Spectra Failure: As the electron spirals inward with continuously varying frequency, it should emit a continuous spectrum of light. In reality, atoms emit sharp, discrete line spectra.

3. Bohr's Postulates for Hydrogen-like Species

In 1913, Niels Bohr combined classical mechanics with Planck's quantum hypothesis to formulate three fundamental postulates for single-electron (hydrogenic) systems (H, He+, Li2+, Be3+):

Postulate 1 (Stationary Non-Radiating Orbits):
Electrons revolve only in certain non-radiating orbits called stationary orbits. In these stable states, the electrostatic Coulomb attraction provides the necessary centripetal force:

m v2r = 14πε0 Z e2r2 = k Z e2r2

Postulate 2 (Quantization of Orbital Angular Momentum):
An electron can only orbit in states where its orbital angular momentum L is an integral multiple of h2π:

L = m v r = n h2π = n ℏ  (where n = 1, 2, 3... is the Principal Quantum Number)

Postulate 3 (Bohr's Quantum Transition Frequency Condition):
Radiation is emitted or absorbed only when an electron jumps from one stationary state (E2) to another (E1):

ΔE = E2 − E1 = h ν = h cλ
Bohr Atomic Model, Postulates, and Energy Level Quantization
Figure 2: Bohr's Model showing Stationary Orbits, Quantized Angular Momentum, and De Broglie Standing Waves

4. Quantitative Derivations of Bohr's Model

A. Orbital Radius (rn)

Eliminating v between Postulate 1 and Postulate 2 yields:

Bohr Orbital Radius Formula
rn = ε0 h2π m e2 n2Z = r1 n2Z = 0.529 × n2Z Å
Proportionality: rn ∝ n2Z  (For Hydrogen Z = 1: r1 = 0.529 Å, r2 = 4r1, r3 = 9r1)

B. Orbital Speed (vn)

Bohr Orbital Speed Formula
vn = e22 ε0 h Zn = 2.18 × 106 × Zn m/s = c137 Zn
Proportionality: vn ∝ Zn  (Fine structure constant α = e22ε0 h c ≈ 1137)

C. Time Period (Tn), Frequency (fn), Current (In) & Magnetic Field (Bn)

Physical ParameterMathematical ExpressionNEET Scaling Proportionality
Time Period (Tn)Tn = 2π rnvn ∝ n2/ZZ/nTn ∝ n3Z2
Orbital Frequency (fn)fn = 1Tn = ω2πfn ∝ Z2n3
Equivalent Current (In)In = eTn = e × fnIn ∝ Z2n3
Magnetic Field at Center (Bn)Bn = μ0 In2 rn ∝ Z2/n3n2/ZBn ∝ Z3n5
Magnetic Dipole Moment (μL)Mn = In × A = In × (π rn2) = n μBMn ∝ n (Bohr Magneton μB = e ℏ2 m)

D. Energy Relations & The Virial Theorem

The total energy of an electron in a stable orbit is the sum of its Kinetic Energy (KE) and Electrostatic Potential Energy (PE):

  • Kinetic Energy (KE): KE = 12 m v2 = k Z e22 r = + 13.6 Z2n2 eV (always positive)
  • Potential Energy (PE): PE = − k Z e2r = − 27.2 Z2n2 eV (negative due to attraction)
  • Total Energy (TE or En): En = KE + PE = − 13.6 Z2n2 eV
The Golden Energy Matrix (Virial Theorem)
TE = − KE = PE2  ⇔  PE = 2 × TE = − 2 × KE
NEET Trap: The negative sign of Total Energy indicates that the electron is bound to the nucleus. If the potential energy reference at infinity is changed from 0 to +10 eV, Kinetic Energy remains completely unchanged, but both PE and Total Energy increase by +10 eV!

5. Hydrogen Spectrum & Spectral Series

When an electron in an excited state n2 drops to a lower state n1, a single photon is emitted with energy:

Rydberg Formula for Emitted Wavelength
1λ = νc = R Z2 [ 1n12 − 1n22 ]
Rydberg Constant: R = m e48 ε02 c h3 ≈ 1.097 × 107 m−1  ⇒  1R ≈ 912 Å = 91.2 nm
Hydrogen Spectrum Energy Level Diagram and Transitions
Figure 3: Hydrogen Spectral Transitions (Lyman, Balmer, Paschen, Brackett, Pfund) and Energy Levels

Comprehensive Master Table of Spectral Series

Seriesn1n2Spectral RegionShortest λ (Series Limit n2 = ∞)Longest λ (First Line n2 = n1 + 1)Ratio λmaxλmin
Lyman12, 3, 4...Ultraviolet (UV)1R ≈ 912 Å43R ≈ 1216 Å43 ≈ 1.33
Balmer23, 4, 5...Visible / Near UV4R ≈ 3646 Å365R ≈ 6563 Å95 = 1.80
Paschen34, 5, 6...Near Infrared9R ≈ 8208 Å1447R ≈ 18751 Å167 ≈ 2.29
Brackett45, 6, 7...Mid Infrared16R ≈ 14592 Å4009R ≈ 40510 Å259 ≈ 2.78
Pfund56, 7, 8...Far Infrared25R ≈ 22800 Å90011R ≈ 74580 Å3611 ≈ 3.27
Rydberg Formula Master Matrix and Transition Limits
Figure 4: Rydberg Wavelength Limits, Number of Spectral Lines, and Virial Energy Partition

Calculation of Number of Spectral Lines

  • For a sample of gas containing many atoms with electrons excited to state n:
    Total Emitted Spectral Lines from State n to Ground
    Ntotal = nC2 = n (n − 1)2
  • Between two specified states n2 and n1 (where n2 > n1):
    Spectral Lines between Level n2 and n1
    N = Δn (Δn + 1)2  (where Δn = n2 − n1)
  • CRITICAL NEET DISTINCTION (Single Isolated Atom): In a single isolated atom containing one electron excited to state n, the electron can take only one cascade path downwards. Thus, the maximum number of lines emitted is (n − 1), not n(n−1)2!

6. De Broglie's Explanation of Bohr's 2nd Postulate

In 1923, Louis de Broglie showed that an electron revolving in a stable Bohr orbit behaves as a circular stationary (standing) matter wave.

  • For constructive interference of the matter wave around a circular orbit of circumference 2πr, the circumference must contain an integral number of de Broglie wavelengths:
    Standing Wave Condition
    2 π rn = n λ
  • Substituting de Broglie's wavelength λ = hp = hm v:
    2 π rn = n hm v  ⇒  m v rn = n h2π
  • Conclusion: Bohr's arbitrary quantization assumption is a direct, natural consequence of electron wave-particle duality!

7. Atomic Excitation, Ionization & Recoil Dynamics

Definitions & Key Benchmarks for Hydrogen

  • Ground State: n = 1, E1 = − 13.60 eV.
  • First Excited State: n = 2, E2 = − 3.40 eV. Excitation energy ΔE = 10.20 eV.
  • Second Excited State: n = 3, E3 = − 1.51 eV. Excitation energy ΔE = 12.09 eV.
  • Third Excited State: n = 4, E4 = − 0.85 eV. Excitation energy ΔE = 12.75 eV.
  • Ionization Energy (IE): The minimum energy required to liberate an electron from ground state to infinity: IE = E∞ − E1 = 0 − (− 13.6) = + 13.60 eV (for H). For hydrogen-like ion: 13.6 Z2 eV.
  • Ionization Potential (IP): IP = 13.6 Z2 Volts.

Recoil of Emitting Atom

When an isolated stationary atom of mass M emits a photon of frequency ν and wavelength λ, linear momentum is strictly conserved:

Recoil Momentum & Velocity of Atom
precoil = pphoton = hλ = ΔEc
vrecoil = precoilM = hλ M = ΔEM c
Krecoil = p22 M = (ΔE)22 M c2

8. Limitations of Bohr's Atomic Model

Key Failures Tested in NEET:
  1. Applies Only to Single-Electron Systems: Accurately predicts spectra for H, He+, Li2+, Be3+, but completely fails for multi-electron atoms (even neutral Helium with 2 electrons) due to electron-electron Coulomb repulsion.
  2. Unable to Explain Fine Structure: High-resolution spectroscopy reveals that individual spectral lines are actually doublets or multiplets. Bohr model cannot account for this (explained later by spin-orbit coupling and relativistic corrections).
  3. Zeeman & Stark Effects: Fails to explain the splitting of spectral lines in external magnetic fields (Zeeman effect) or external electric fields (Stark effect).
  4. Relative Intensities: Cannot predict why certain spectral transitions are much brighter (higher transition probability) than others.
  5. Heisenberg Uncertainty Principle Violation: Bohr's model assumes a well-defined circular orbit where both the position and momentum of the electron are known simultaneously with 100% precision, directly violating Δx · Δp ≥ ℏ2.
NEET Master Formulas & 20 Golden Facts - Atoms

NEET Master Formula Matrix & 20 Golden Facts

Chapter: Atoms (Modern Physics) • Complete Revision Sheet for Instant Speed & Precision

Master Formula Matrix: Bohr Model & Atomic Physics

CategoryFormula NameMathematical ExpressionNEET Scaling Proportionality
Rutherford ScatteringDistance of Closest Approachr0 = 14πε0 2 Z e2Ek = 4 k Z e2m v2r0 ∝ ZEk ∝ 1v2
Rutherford ScatteringImpact Parameterb = 14πε0 Z e2 cot(θ2)Ekb ∝ cot(θ2)
Rutherford ScatteringAngular DistributionN(θ) ∝ 1sin4(θ2)N(θ1)N(θ2) = sin4(θ2/2)sin4(θ1/2)
Bohr PostulatesAngular Momentum QuantizationL = m v r = n h2π = n ℏL ∝ n (Independent of Z)
Bohr OrbitOrbital Radiusrn = ε0 h2π m e2 n2Z = 0.529 n2Z Årn ∝ n2Z
Bohr OrbitOrbital Speedvn = e22ε0 h Zn = 2.18 × 106 Zn m/s = c137 Znvn ∝ Zn
Bohr DynamicsOrbital Time PeriodTn = 2π rnvnTn ∝ n3Z2
Bohr DynamicsOrbital Frequencyfn = 1Tn = vn2π rnfn ∝ Z2n3
ElectrodynamicsEquivalent Circular CurrentIn = eTn = e × fnIn ∝ Z2n3
ElectrodynamicsMagnetic Field at NucleusBn = μ0 In2 rnBn ∝ Z3n5
Energy PartitionKinetic EnergyKE = 12 m v2 = + 13.6 Z2n2 eVKE > 0 (always positive)
Energy PartitionPotential EnergyPE = − k Z e2r = − 27.2 Z2n2 eVPE = − 2 × KE
Energy PartitionTotal EnergyEn = KE + PE = − 13.6 Z2n2 eVEn = − KE = PE2
SpectroscopyRydberg Transition Formula1λ = R Z2 [1n12 − 1n22]R ≈ 1.097 × 107 m−1, 1R ≈ 912 Å
Spectral LinesTotal Lines from State n to GroundNtotal = n (n − 1)2Gas sample containing many atoms
Spectral LinesTransitions between n2 and n1N = Δn (Δn + 1)2where Δn = n2 − n1
De Broglie WavesStanding Wave Condition2π rn = n λn-th orbit has exactly n de Broglie waves
Recoil MechanicsRecoil Speed of Emitting Atomvrecoil = hλ M = ΔEM cConservation of Linear Momentum

20 Golden NEET Must-Know Facts & Traps

Golden Fact 01 Closest Approach Scaling
Distance of closest approach scales inversely with kinetic energy and square of velocity: r0 ∝ ZEk ∝ 1v2. Doubling the velocity cuts r0 by a factor of 4.
Golden Fact 02 NEET Trap: Impact Parameter
Head-on collision means b = 0 ⇒ θ = 180° (direct rebound). An impact parameter b → ∞ ⇒ θ → 0° (undeflected). Never confuse b = 0 with zero deflection!
Golden Fact 03 Rutherford Angular Ratio
In Rutherford scattering, the number of particles scattered at angle θ obeys: N(θ) ∝ cosec4(θ2). For example, N(60°)N(90°) = sin4(45°)sin4(30°) = (1/√2)4(1/2)4 = 1/41/16 = 4.
Golden Fact 04 Atom vs Nuclear Volume Ratio
Atomic radius is ∼ 10−10 m (1 Å); Nuclear radius is ∼ 10−15 m (1 fm). The volume of the atom is VatomVnucleus ∼ 1015 times larger than the nucleus! The atom is mostly hollow space.
Golden Fact 05 Bohr Radius Ratios
Radius of n-th orbit: rn = 0.529 × n2Z Å. For Hydrogen (Z = 1), the radii are in the ratio r1 : r2 : r3 : r4 = 1 : 4 : 9 : 16.
Golden Fact 06 Electron Speed in First Orbit
In the first Bohr orbit of Hydrogen (n = 1, Z = 1), electron speed is v1 = c137 ≈ 2.18 × 106 m/s. As n increases, speed decreases inversely: vn ∝ 1 / n.
Golden Fact 07 Time Period & Frequency Trap
Time period scales as Tn ∝ n3Z2, NOT n2! Orbital frequency scales as fn ∝ Z2n3. The ratio of time periods for n = 1 to n = 2 in Hydrogen is 1 : 8.
Golden Fact 08 Magnetic Field at Center Scaling
Magnetic field produced at the nucleus by the revolving electron: Bn ∝ Z3n5. Frequent NEET question! Ratio B1 / B2 for Hydrogen is 25 / 15 = 32.
Golden Fact 09 Virial Theorem Super-Rule
For any Coulomb bound system: TE = − KE = PE2  ⇒  PE = 2 × TE = − 2 × KE. If TE = − 3.4 eV, then KE = + 3.4 eV, and PE = − 6.8 eV instantly!
Golden Fact 10 Must-Memorize Energy Levels (H-Atom)
Memorize the first 4 levels of Hydrogen cold: E1 = − 13.60 eV, E2 = − 3.40 eV, E3 = − 1.51 eV, E4 = − 0.85 eV, E∞ = 0 eV.
Golden Fact 11 Crucial Excitation Energies
First excitation energy (1 → 2): 10.20 eV.
Second excitation energy (1 → 3): 12.09 eV.
Third excitation energy (1 → 4): 12.75 eV.
If an incident electron has 11.5 eV, it can only excite H to n = 2 and leaves with 1.3 eV residual KE!
Golden Fact 12 Ionization Energy vs Potential
Ionization energy is measured in electron-volts (eV): IE = 13.6 Z2 eV.
Ionization potential is measured in Volts (V): IP = 13.6 Z2 V.
For He+ (Z = 2): IE = 54.4 eV, IP = 54.4 V. For Li2+ (Z = 3): IE = 122.4 eV, IP = 122.4 V.
Golden Fact 13 Rydberg Reciprocal Constant
Never divide by 1.097 × 107 during exam calculations! Use the golden shortcut: 1R ≈ 912 Å = 91.2 nm. For Lyman limit: λ = 1/R = 912 Å. For Balmer limit: λ = 4/R = 4 × 912 = 3648 Å.
Golden Fact 14 Visible Region Exclusivity
ONLY the Balmer series produces lines in the visible region of the electromagnetic spectrum! Lyman is strictly Ultraviolet. Paschen, Brackett, and Pfund are strictly Infrared.
Golden Fact 15 Series Limit Definition
Series Limit always corresponds to transition from n2 = ∞ to n1.
This represents the maximum energy photon, maximum frequency, and minimum wavelength (λmin) in that series!
Golden Fact 16 First Line / Longest Wavelength
The first line of any series corresponds to transition from n2 = n1 + 1 to n1.
This represents the minimum energy photon, minimum frequency, and longest wavelength (λmax) in that series!
Golden Fact 17 Balmer Wavelength Extreme Ratio
For Balmer series: λmax = 365R (3 → 2), λmin = 4R (∞ → 2).
The ratio is: λmaxλmin = 36/5R4/R = 95 = 1.80.
Golden Fact 18 Single Isolated Atom Trap!
In a gas sample with millions of atoms excited to state n, total spectral lines = n(n−1)2.
In a single isolated atom, the electron takes a single path down: Maximum lines = (n − 1)! NEET loves this trap.
Golden Fact 19 De Broglie Standing Wave Theorem
The n-th Bohr orbit contains exactly n de Broglie wavelengths: 2π rn = n λ. For example, the 4th orbit of Hydrogen has exactly 4 complete matter waves around its circumference.
Golden Fact 20 Recoil Speed of Emitting Atom
When a stationary atom of mass M emits a photon of energy ΔE, recoil speed is: vrecoil = ΔEM c. The fraction of energy lost to recoil kinetic energy is minuscule (∼ 10−8), so photon energy ≈ ΔE.
NEET Physics: Atoms - 8 Solved Numericals (3-Box Pedagogy)

NEET Physics: Atoms — 8 Solved Numericals

Master Problem Solving with Strict 3-Box Pedagogy: Given & Concept • Step-by-Step Calculation • NEET Speed Shortcut

Numerical 01: Distance of Closest Approach & Energy Scaling
An α-particle of kinetic energy 5.0 MeV is projected directly towards a stationary gold nucleus (Z = 79). Calculate the distance of closest approach r0. If the kinetic energy of the α-particle is doubled, what will be the new distance of closest approach? (Take 14πε0 = 9 × 109 N·m2/C2, e = 1.6 × 10−19 C).
Box 1: Given Data & Core Concept

• Given: Kinetic energy Ek = 5.0 MeV = 5.0 × 106 × 1.6 × 10−19 J = 8.0 × 10−13 J.
• Target nucleus: Gold (Z = 79), α-particle charge q1 = 2e, target nucleus charge q2 = Ze = 79e.
• Core Concept: At the distance of closest approach r0 (turning point), entire KE is converted into electrostatic potential energy U:
Ek = 14πε0 (2e)(Ze)r0 ⇒ r0 = 14πε0 2 Z e2Ek.

Box 2: Step-by-Step Mathematical Calculation
1. Substitute the values into the closest approach equation:
r0 = (9 × 109) × 2 × 79 × (1.6 × 10−19)2 8.0 × 10−13
2. Simplifying numerator: (9 × 109) × 158 × (2.56 × 10−38) = 3.64 × 10−27.
3. Solving for r0:
r0 = 3.64 × 10−27 8.0 × 10−13 = 4.55 × 10−14 m = 45.5 fm
4. When kinetic energy is doubled (E'k = 2 Ek):
r'0 = r02 = 45.5 fm2 = 22.75 fm
Final Answer: r0 = 4.55 × 10−14 m (45.5 fm); New r'0 = 2.275 × 10−14 m (22.75 fm).
Box 3: NEET Speed Shortcut & Pro Tip

NEET Mental Shortcut: r0 ∝ ZEk ∝ 1v2.
• If KE is multiplied by factor k, r0 becomes r0 / k.
• If speed v is doubled, r0 becomes r0 / 4!
• For gold (Z = 79) and 5 MeV: r0 ≈ 4.5 × 10−14 m. Always verify order of magnitude is ∼ 10−14 m (10 to 50 fm).

Numerical 02: Bohr Orbit Radius & Electron Speed in Li2+
Calculate the radius of the third excited state (n = 4) of a doubly ionized Lithium ion (Li2+, Z = 3). Also determine the ratio of the electron's speed in this state to that in the ground state of a neutral Hydrogen atom. (Given: Bohr radius r1,H = 0.529 Å).
Box 1: Given Data & Core Concept

• Given: Li2+ is a hydrogen-like species with Z = 3.
• "Third excited state" means principal quantum number n = 3 + 1 = 4.
• Core Concepts:
1. Orbital Radius: rn = r1,H × n2Z.
2. Orbital Speed: vn = v1,H × Zn.

Box 2: Step-by-Step Mathematical Calculation
1. Calculate radius for n = 4 and Z = 3:
r4(Li2+) = 0.529 × 423 = 0.529 × 163 = 0.529 × 5.333 = 2.821 Å
2. Calculate the velocity ratio:
v(Li2+, n=4)v(H, n=1) = Z / n1 / 1 = 34 = 0.75
Final Answer: Radius r4 = 2.82 Å; Speed ratio = 3 : 4 = 0.75.
Box 3: NEET Speed Shortcut & Pro Tip

NEET Vocabulary Warning:
• "1st excited state" ⇒ n = 2
• "2nd excited state" ⇒ n = 3
• "m-th excited state" ⇒ n = m + 1!
Students commonly plug n = 3 instead of n = 4 and get negative marks. Always write n = excited state + 1 first!

Numerical 03: Wavelength & Frequency of Emitted Photon (3 → 1 Transition)
In a hydrogen atom, an electron transitions from the second excited state (n = 3) to the ground state (n = 1). Calculate the wavelength (λ) and frequency (ν) of the emitted radiation. (Given: Rydberg constant R = 1.097 × 107 m−1, c = 3.0 × 108 m/s).
Box 1: Given Data & Core Concept

• Given: Transition n2 = 3 → n1 = 1 (Lyman series, β-line). Hydrogen atom (Z = 1).
• Core Concept: Rydberg Transition Formula:
1λ = R Z2 [ 1n12 − 1n22 ]  and  ν = cλ.

Box 2: Step-by-Step Mathematical Calculation
1. Substitute quantum numbers:
1λ = R [ 112 − 132 ] = R [ 1 − 19 ] = 8 R9
2. Invert to solve for λ using 1R ≈ 912 Å:
λ = 98 R = 98 × 912 Å = 9 × 114 Å = 1026 Å = 102.6 nm
3. Frequency ν = cλ:
ν = 3.0 × 1081.026 × 10−7 = 2.92 × 1015 Hz
Final Answer: λ = 102.6 nm (1026 Å in UV region); ν = 2.92 × 1015 Hz.
Box 3: NEET Speed Shortcut & Pro Tip

Energy Difference Shortcut:
E1 = − 13.60 eV, E3 = − 1.51 eV.
Photon energy: ΔE = 13.60 − 1.51 = 12.09 eV.
Using the ultra-fast formula: λ (Å) = 12400ΔE (eV) = 1240012.09 ≈ 1026 Å! Solved in 5 seconds flat!

Numerical 04: Balmer Series Extreme Wavelength Ratio
Find the ratio of the longest wavelength (λmax) to the shortest wavelength (λmin) in the Balmer series of a Hydrogen atom spectrum.
Box 1: Given Data & Core Concept

• Balmer Series: Lower state n1 = 2.
• Longest wavelength (λmax): Minimum energy transition, i.e., from adjacent upper level n2 = n1 + 1 = 3 (Hα line).
• Shortest wavelength (λmin): Maximum energy transition (series limit), i.e., from n2 = ∞.

Box 2: Step-by-Step Mathematical Calculation
1. For longest wavelength (3 → 2):
1λmax = R [ 122 − 132 ] = R [ 14 − 19 ] = 5 R36  ⇒  λmax = 365 R
2. For shortest wavelength (∞ → 2):
1λmin = R [ 122 − 1∞2 ] = R4  ⇒  λmin = 4R
3. Taking the ratio:
λmaxλmin = 36 / 5R4 / R = 365 × 4 = 95 = 1.80
Final Answer: λmaxλmin = 95 = 1.80.
Box 3: NEET Speed Shortcut & Pro Tip

Universal Series Ratio Formula:
For any spectral series with base level n1:
λmaxλmin = (n1 + 1)22 n1 + 1.
• Lyman (n1 = 1): (2)2 / 3 = 4 / 3.
• Balmer (n1 = 2): (3)2 / 5 = 9 / 5.
• Paschen (n1 = 3): (4)2 / 7 = 16 / 7. Never recalculate in NEET!

Numerical 05: Magnetic Field at the Nucleus in Bohr Orbits
The magnetic field at the center of a hydrogen atom in the ground state is B1. Find the magnetic field at the center when the electron is excited to the second excited state (n = 3) in terms of B1.
Box 1: Given Data & Core Concept

• Given: Initial state n = 1 (field B1); Final state: 2nd excited state ⇒ n = 3.
• Core Concept: Revolving electron constitutes an electric current loop:
I = eT ∝ Z2n3.
Orbital radius r ∝ n2Z.
Magnetic field at center: B = μ0 I2 r ∝ Z2/n3n2/Z = Z3n5.

Box 2: Step-by-Step Mathematical Calculation
1. Formulate proportionality for Hydrogen (Z = 1):
Bn ∝ 1n5
2. Set up ratio for n = 3 vs n = 1:
B3B1 = 1535 = 1243
3. Solving for B3:
B3 = B1243
Final Answer: B3 = B1243 ≈ 4.12 × 10−3 B1.
Box 3: NEET Speed Shortcut & Pro Tip

Power of n Trap:
• Radius: r ∝ n2
• Velocity: v ∝ 1/n
• Time Period: T ∝ n3
• Current: I ∝ 1/n3
• Magnetic Field: B ∝ 1/n5 (NEET examiners test this 5th power frequently!).

Numerical 06: Photon Excitation, Total Spectral Lines & Visible Lines
A sample of hydrogen atoms in their ground state is irradiated with photons of energy 12.75 eV. Determine: (a) the principal quantum number of the state to which electrons are excited, (b) the total number of spectral lines emitted upon de-excitation, and (c) how many of these lines fall in the visible region.
Box 1: Given Data & Core Concept

• Given: Incident photon energy Ephoton = 12.75 eV. Initial state n = 1 (E1 = − 13.60 eV).
• Energy of excited state: En = E1 + Ephoton = − 13.60 + 12.75 = − 0.85 eV.
• Total lines in gas sample: Ntotal = n(n − 1)2.
• Visible lines belong strictly to the Balmer series (transitions ending at n = 2).

Box 2: Step-by-Step Mathematical Calculation
1. Find principal quantum number n:
En = − 13.6n2 = − 0.85 eV  ⇒  n2 = 13.60.85 = 16  ⇒  n = 4
2. Calculate total emitted lines from n = 4:
Ntotal = 4 × (4 − 1)2 = 122 = 6 lines
3. Break down by series:
• Lyman (transitions to n = 1): (4 → 1, 3 → 1, 2 → 1) = 3 lines (UV)
• Balmer (transitions to n = 2): (4 → 2, 3 → 2) = 2 lines (Visible)
• Paschen (transitions to n = 3): (4 → 3) = 1 line (IR)
Total = 3 + 2 + 1 = 6 lines.
Final Answer: Excited state n = 4; Total spectral lines = 6; Visible lines = 2 (from Balmer series).
Box 3: NEET Speed Shortcut & Pro Tip

Instant Breakdown Formula:
When de-exciting from level n to ground state:
• Number of Lyman lines = n − 1
• Number of Balmer lines = n − 2
• Number of Paschen lines = n − 3
For n = 4: Lyman = 3, Balmer = 2, Paschen = 1. Immediate answer without drawing!

Numerical 07: Recoil Speed & Recoil Energy of Hydrogen Atom
A stationary hydrogen atom in its first excited state (n = 2) de-excites to the ground state (n = 1) by emitting a photon. Calculate the recoil speed of the hydrogen atom and the kinetic energy of recoil. (Mass of hydrogen atom M = 1.67 × 10−27 kg, c = 3.0 × 108 m/s).
Box 1: Given Data & Core Concept

• Given: Transition 2 → 1. ΔE = 10.20 eV = 10.20 × 1.6 × 10−19 J = 1.632 × 10−18 J.
• Mass M = 1.67 × 10−27 kg.
• Core Concept: Linear momentum conservation: Initial momentum = 0 ⇒ precoil = pphoton = ΔEc.
• Recoil speed: vrecoil = precoilM = ΔEM c.
• Recoil kinetic energy: Krecoil = p22 M.

Box 2: Step-by-Step Mathematical Calculation
1. Recoil momentum p:
p = 1.632 × 10−183.0 × 108 = 5.44 × 10−27 kg·m/s
2. Recoil speed vrecoil:
vrecoil = 5.44 × 10−271.67 × 10−27 = 3.26 m/s
3. Recoil kinetic energy:
Krecoil = (5.44 × 10−27)22 × (1.67 × 10−27) = 8.86 × 10−27 J ≈ 5.54 × 10−8 eV
Final Answer: Recoil speed = 3.26 m/s; Recoil KE = 5.54 × 10−8 eV (negligible compared to 10.2 eV).
Box 3: NEET Speed Shortcut & Pro Tip

Conceptual Takeaway:
The recoil kinetic energy is only ~ 10−8 eV, which is less than 1 part in 100 million of the photon's energy! That is why in NEET calculations, we can safely treat the emitted photon energy as virtually identical to the atomic transition energy (ΔE).

Numerical 08: Electron-Atom Collision & Inelastic Scattering Threshold
A beam of monoenergetic electrons collides with ground-state hydrogen atoms. Determine: (a) the minimum kinetic energy of the electrons required to excite the Lyman-α line, (b) the minimum kinetic energy required to excite the Balmer-α line, and (c) if incident electrons have KE = 11.0 eV, what type of collision occurs and what is the kinetic energy of the scattered electrons?
Box 1: Given Data & Core Concept

• Lyman-α line: Transition 2 → 1. Requires excitation from n = 1 to n = 2 (ΔE1→2 = 10.20 eV).
• Balmer-α line: Transition 3 → 2. Requires excitation from n = 1 to at least n = 3 (ΔE1→3 = 12.09 eV).
• Collision Rules for Electrons: Because the electron's mass is tiny compared to the nucleus, the electron can transfer almost any required quantized excitation energy ΔE to the bound electron. If Eincident < ΔE1→2 (10.2 eV), collision is purely elastic. If Eincident ≥ 10.2 eV, inelastic collision can occur.

Box 2: Step-by-Step Mathematical Calculation
1. Minimum KE for Lyman-α:
Emin(Lyman-α) = E2 − E1 = − 3.40 − (− 13.60) = 10.20 eV
2. Minimum KE for Balmer-α: To emit Balmer-α (3 → 2), the atom must first be raised to n = 3:
Emin(Balmer-α) = E3 − E1 = − 1.51 − (− 13.60) = 12.09 eV
3. For incident electron with KE = 11.0 eV: Since 10.20 eV ≤ 11.0 eV < 12.09 eV, the electron can excite the atom to n = 2 only (inelastic collision):
KEscattered = Eincident − ΔE1→2 = 11.00 − 10.20 = 0.80 eV
(If collision is elastic, scattered KE remains 11.0 eV).
Final Answer: (a) 10.20 eV; (b) 12.09 eV; (c) Inelastic collision occurs, scattered electron has KE = 0.80 eV (or 11.0 eV if elastic).
Box 3: NEET Speed Shortcut & Pro Tip

Photon vs Electron Collision Trap:
• A photon must match ΔE EXACTLY (resonant absorption). A photon of 11.0 eV CANNOT be absorbed by ground state hydrogen because no energy level exists at −2.6 eV!
• An electron does NOT need exact match; it can impart 10.2 eV and keep the remaining 0.8 eV as kinetic energy!

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