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Atoms

Atoms

Physics · Class 12

1. Rutherford Scattering Experiment

A beam of α-particles (He²⁺, charge +2e) is fired at a very thin gold foil; scattered α-particles are detected on a ZnS screen placed around the foil.

Observations & Conclusions :

Most α-particles pass undeflected ⇒ most of the atom is empty space; the nucleus occupies a very small volume.
A few are deflected by large angles, very few (≈ 1 in 8000) bounce back (θ > 90°) ⇒ the positive charge and almost all the mass is concentrated in a tiny central nucleus, which repels the positive α-particles.
Nuclear model : electrons revolve around this nucleus.

2. Distance of Closest Approach

Head-on collision (b = 0) : α-particle with kinetic energy K stops momentarily at distance r₀; all KE → electric PE.

K = 14πε₀ · (2e)(Ze)r₀

r₀ = 14πε₀ · 2Ze²K

e.g. 7.7 MeV α-particles on gold (Z = 79) :
r₀ = (9×10⁹)(2×79)(1.6×10⁻¹⁹)²7.7×10⁶ × 1.6×10⁻¹⁹
r₀ ≈ 3.0 × 10⁻¹⁴ m ≈ 30 fm ⇒ size of nucleus is smaller than this.

3. Impact Parameter (b)

b = perpendicular distance of the initial velocity line of the α-particle from the centre of the nucleus.

b = 14πε₀ · Ze² cot(θ/2)K

θ = scattering angle, K = ½mv² of the α-particle.
Small b ⇒ large θ. b = 0 (head-on) ⇒ θ = 180° (bounces back).
Large b ⇒ θ ≈ 0 (passes almost undeflected).

4. Bohr's Atomic Model

Electrons revolve around the nucleus like planets around the sun.
Bohr gave three postulates for single-electron (hydrogen-like) atoms : H (Z = 1), He⁺ (Z = 2), Li²⁺ (Z = 3).

5. 1^st Postulate

The electron gets the centripetal force from the electrostatic attraction of the nucleus.

F = mv²r = 14πε₀ · (Ze)(e)r²

mv²r = Ze²4πε₀ ...(i)

Key Points :

Rutherford : tiny, dense, positive nucleus; atom mostly empty.
r₀ ∝ Z/K ; b ∝ cot(θ/2)/K.
Bohr's model is valid only for one-electron atoms/ions.

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Bohr's Model

6. 2^nd Postulate

Electron revolves only in certain definite circular paths called stable orbits (shells).
In a stable orbit the electron does not emit electromagnetic radiation (contrary to Maxwell's electromagnetism).
In these orbits, angular momentum of the electron about the nucleus is an integral multiple of h/2π (Bohr's quantization) :

L = mvr = nh2π ...(ii) n = 1, 2, 3, ...

7. 3^rd Postulate

Electron emits or absorbs energy only when it makes a transition from one stable orbit to another.
Energy of the photon absorbed / emitted = energy difference between the orbits.

hν = E_n₂ − E_n₁ (energy of photon = energy gap)

8. Speed (v_n) and Radius (r_n)

(i) ÷ (ii) : v = Ze²/4πε₀nh/2π

v_n = Ze²2nhε₀ = v₀Zn ; v₀ ≈ 2.18 × 10⁶ m/s

From (ii) : r = nh2πmv = nh2πm · 2nhε₀Ze²

r_n = n²h²ε₀πmZe² = a₀n²Z ; a₀ ≈ 0.529 Å

a₀ (= r₀) = first Bohr radius of H; v₀ ≈ c/137 = speed in first orbit of H.

9. Energy of n^th Orbit

Kinetic energy :

K_n = ½mv² = ½m(Ze²2nhε₀)²

K_n = mZ²e⁴8n²h²ε₀²

Electric potential energy :

U_n = 14πε₀ · (Ze)(−e)r_n = −Ze²4πε₀ · πmZe²n²h²ε₀

U_n = −mZ²e⁴4n²h²ε₀²

Total energy : E_n = K_n + U_n

E_n = −mZ²e⁴8n²h²ε₀² = −(13.6 eV)Z²n²

Negative energy ⇒ electron is bound to the nucleus. E = 0 at n = ∞ (free electron at rest).

10. Bohr's Model – Summary

Quantity	Formula	varies as
Orbit radius	r_n = (0.529 Å) n²/Z	n²/Z
Orbital speed	v_n = (2.18×10⁶ m/s) Z/n	Z/n
Total energy	E_n = −(13.6 eV) Z²/n²	Z²/n²
Kinetic energy	K_n = −E_n	Z²/n²
Potential energy	U_n = 2E_n	Z²/n²

E₀ = 13.6 eV = Rhc = Rydberg unit of energy.
K : U : E = 1 : −2 : −1

11. Hydrogen-like Atoms

Atom	Z	r₁	v₁ (m/s)	E₁
H	1	0.529 Å	2.18×10⁶	−13.6 eV
He⁺	2	0.265 Å	4.36×10⁶	−54.4 eV
Li²⁺	3	0.176 Å	6.54×10⁶	−122.4 eV

(E_n)_He⁺ = 4(E_n)_H ; (ΔE)_He⁺ = 4(ΔE)_H

(E_n)_Li²⁺ = 9(E_n)_H ; (ΔE)_Li²⁺ = 9(ΔE)_H

e.g. He⁺ : n = 1 → 2 needs 4 × 10.2 = 40.8 eV; E₁ = −4(13.6) eV, E₂ = −4(3.4) eV.

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Energy Levels & Spectra

12. Hydrogen Atom Energy Levels (Z = 1)

E_n = −13.6n² eV

Energy gaps (eV) : 1→2 = 10.2 ; 1→3 = 12.09 ; 1→4 = 12.75 ; 2→3 = 1.89 ; 2→4 = 2.55 ; 3→4 = 0.66.
Levels come closer as n increases. E₇ = −0.28 eV, E₈ = −0.21 eV.

13. Energy Gap (ΔE)

ΔE = |E_n₁ − E_n₂| with E_n = −(13.6 eV)Z²/n²

ΔE = (13.6 eV) Z² |1n₁² − 1n₂²|

14. Emission & Absorption Spectra

White (polychromatic) light through a prism gives a continuous spectrum.
Pass white light through hydrogen gas : H atoms absorb only those wavelengths whose photon energy equals an energy gap ⇒ dark lines on a continuous background = absorption spectrum.
The excited atoms then de-excite and emit light in all directions; seen through a prism this gives bright lines on a dark background = emission (line) spectrum.
Absorption lines and emission lines of a gas occur at the same wavelengths — a fingerprint of the element.

15. Excitation of Electron (by Photons)

An electron in a lower orbit absorbs a photon only if photon energy = an energy gap between orbits, or photon energy ≥ ionization energy.
A photon is absorbed completely or not at all.

e.g. H atom in ground state; which photons are absorbed?
2, 3, 5, 8, 10 eV → ✗ ; 10.2 eV → ✓ (n = 1→2)
10.5, 11, 12 eV → ✗ ; 12.1 eV → ✓ (1→3)
12.5 eV → ✗ ; 12.75 eV → ✓ (1→4)
13.6, 13.7, 14, 14.5, 15 eV → ✓ (≥ 13.6 eV, ionization)

If E > 13.6 eV : electron is ejected (free electron)

K.E. of free e⁻ = E − 13.6 eV

e.g. E = 15 eV ⇒ K.E. = 15 − 13.6 = 1.4 eV

16. Photons Don't Add Up

Beam of 12.1 eV photons on H gas (n = 1) ⇒ electron goes n = 1 → n = 3.
Beam containing 10.2 eV + 1.9 eV photons ⇒ only the 10.2 eV photon is absorbed (n = 1 → 2). The two photons do not combine to give 12.1 eV; the 1.9 eV photon passes through (an electron already in n = 2 could absorb it, taking it 2 → 3).

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Transitions & Series

17. De-excitation of Electron

The electron stays in an excited state for a very short time (~10⁻⁸ s, i.e. a few ns), then jumps to lower orbits by emitting photons, finally reaching the ground state (n = 1).
It may return directly (4 → 1) or in steps (4 → 3 → 1, 4 → 2 → 1, 4 → 3 → 2 → 1 ...).

18. Number of Spectral Lines

If electron is excited to the n^th state, the number of different wavelengths (lines) in the emission spectrum :

N = ⁿC₂ = n(n − 1)2

n = 4 ⇒ N = 4×3/2 = 6 lines (for a single atom at most n − 1 lines are emitted at a time).

19. Energy & Wavelength of Photon

E = (13.6 eV)Z²|1n₁² − 1n₂²| = hν = hcλ

λ (in Å) = 12400 eV·ÅE (in eV)

1λ = RZ²(1n₁² − 1n₂²) ⇒ λ = 912 ÅZ²(1/n₁² − 1/n₂²)

Rydberg constant R = 1.097 × 10⁷ m⁻¹ ; 1/R ≈ 912 Å.
E = hc/λ = (hcR)Z²(1/n₁² − 1/n₂²) ⇒ hcR = 13.6 eV (Rydberg unit of energy).
Don't confuse : E_n = energy of electron in n^th orbit (negative); E = energy of photon = energy gap (positive).

20. Spectral Series of Hydrogen

Series	n₁	E (eV)	λ (nm)	Region
Lyman	1	10.2–13.6	122–91	UV
Balmer	2	1.89–3.4	656–365	Visible (+ near UV)
Paschen	3	0.66–1.51	1875–821	IR
Brackett	4	0.31–0.85	4051–1459	IR
Pfund	5	0.17–0.544	7458–2280	Far IR

Series limit (n₂ = ∞) → max^m E, min^m λ : λ_min = 912 n₁² Å.
First line (n₂ = n₁ + 1) → min^m E, max^m λ.
Lyman : E_max = 13.6 eV, λ_min = 12400/13.6 ≈ 912 Å ; E_min = 10.2 eV.
Balmer : E_max = 3.4 eV, λ_min = 12400/3.4 ≈ 3647 Å ; E_min = 1.89 eV (Hα, 656 nm, red).
Visible light ≈ 1.77 eV – 3.1 eV (700 nm – 400 nm) ⇒ the first four Balmer lines (Hα, Hβ, Hγ, Hδ) are visible.

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Important Terms

21. Other Important Terms (H atom)

Ionization energy (I.E.) : min^m energy needed to remove the electron from the ground state (n = 1 → ∞). For H : 13.6 eV.
Ionization potential = I.E./e. For H : 13.6 V.
n^th excitation energy : energy for n = 1 → (n + 1).
1^st (1 → 2) = 10.2 eV ; 2^nd (1 → 3) = 12.09 eV.
n^th excitation potential = excitation energy / e.
1^st E.P. = 10.2 V ; 2^nd E.P. = 12.09 V.
Binding energy of n^th orbit : energy to remove the electron from the n^th orbit = |E_n|.
n = 1 : 13.6 eV ; n = 2 : 3.4 eV ; n = 3 : 1.51 eV.

For hydrogen-like ions multiply by Z² : I.E. of He⁺ = 54.4 eV, of Li²⁺ = 122.4 eV.

22. Other Derived Quantities

Use v = v₀Z/n and r = a₀n²/Z (v₀, a₀ : values for Z = 1, n = 1)

Quantity	Expression	varies as
Area of orbit	A = πr² = (πa₀²) n⁴/Z²	n⁴/Z²
Momentum	p = mv = mv₀(Z/n)	Z/n
Moment of inertia	I = mr² = ma₀² n⁴/Z²	n⁴/Z²
Time period	T = 2πr/v = (2πa₀/v₀) n³/Z²	n³/Z²
Frequency, ω	f = 1/T ; ω = 2π/T = v/r	Z²/n³
Angular momentum	L = mvr = (mv₀a₀) n = nh/2π	n
Equivalent current	i = e/T = ev/(2πr)	Z²/n³
B at centre	B = μ₀i/2r = μ₀ev/(4πr²)	Z³/n⁵
Magnetic moment	M = (e/2m)L = neh/(4πm)	n
de-Broglie λ	λ = h/mv = (3.32 Å) n/Z	n/Z

Area – graph of ln A vs ln n :

ln A = 4 ln n − 2 ln Z + ln A₀ (A₀ = πa₀²)

de-Broglie wavelength :

λ_e = hmv = hmv₀ · nZ = (3.32 Å)nZ

Bohr's condition mvr = nh/2π ⇔ 2πr = nλ : the orbit holds a whole number of de-Broglie waves (standing wave).

Average force & torque (transition between orbits) :

F→_avg = Δp→Δt = m(v→_f − v→_i)Δt , τ→_avg = L→_f − L→_iΔt

Key Points :

Excitation energy is always measured from the ground state (n = 1).
Binding energy of n^th orbit = 13.6 Z²/n² eV.
L and M ∝ n (independent of Z) ; B ∝ Z³/n⁵ ; T ∝ n³/Z².
Ratio M/L = e/2m (gyromagnetic ratio) — same for every orbit.

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Solved Questions

23. Solved Questions

Q1. Assuming the validity of Bohr's model for hydrogen-like ions, the radius of Li⁺⁺ ion in its ground state is 1Xa₀, where X = ? (a₀ = first Bohr radius) [JEE Main April 2025]
(1) 2 (2) 1 (3) 3 (4) 9
Sol. r = a₀n²Z ; n = 1, Z = 3 ⇒ r = a₀3 ⇒ X = 3, option (3)

Q2. An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, find radius, speed and kinetic energy of the n^th orbit.
Sol. Magnetic force qvB (towards centre) provides centripetal force :
evB = mv²r ⇒ mv = eBr ...(i)
mvr = nh2π ...(ii)
(ii) ÷ (i) : r = nh2πeBr ⇒ r_n = √(nh / 2πeB)
v = eBmr ⇒ v_n = eBm√(nh / 2πeB) = √(nheB / 2πm²)
K = ½mv² = ½m · e²B²m² · nh2πeB ⇒ K_n = nheB4πm

Q3. Find the wavelength of the first line of the Balmer series (Hα) of hydrogen.
Sol. n₁ = 2, n₂ = 3 : 1λ = R(14 − 19) = 5R36
λ = 365R = 36 × 9125 Å ≈ 6566 Å ≈ 656 nm (red)
Check : ΔE = 13.6(1/4 − 1/9) = 1.89 eV ⇒ λ = 12400/1.89 ≈ 6560 Å ✓

Q4. Hydrogen atoms in the ground state absorb 12.75 eV photons. Find (a) the excited state, (b) number of emission lines, (c) longest and shortest wavelengths emitted.
Sol. (a) E_n = −13.6 + 12.75 = −0.85 eV = −13.6/n² ⇒ n = 4
(b) N = 4×3/2 = 6 lines
(c) Longest λ : smallest gap 4 → 3 (Paschen) :
λ = 912 Å(1/9 − 1/16) = 912 × 1447 ≈ 18 760 Å ≈ 1.88 μm
Shortest λ : largest gap 4 → 1 (Lyman) :
λ = 912 Å(1 − 1/16) = 912 × 1615 ≈ 973 Å

Q5. Ratio of radius of He⁺ in n = 2 to that of H in n = 1? Also speed ratio.
Sol. r ∝ n²/Z : r_He⁺r_H = 2²/21²/1 = 2 : 1 (r = 1.058 Å)
v ∝ Z/n : v_He⁺v_H = 2/21/1 = 1 : 1

Q6. Ratio of the series-limit wavelengths of Lyman and Balmer series.
Sol. λ_limit = 912 n₁² Å ⇒ λ_L : λ_B = 1² : 2² = 1 : 4 (912 Å : 3648 Å)

24. Quick Formula Revision

Quantity	Formula
Radius	r_n = 0.529 n²/Z Å
Speed	v_n = 2.18×10⁶ Z/n m/s
Energy	E_n = −13.6 Z²/n² eV ; K = −E, U = 2E
Photon	1/λ = RZ²(1/n₁² − 1/n₂²) ; λ(Å) = 12400/E(eV)
No. of lines	n(n − 1)/2
Closest approach	r₀ = 2kZe²/K
Impact parameter	b = kZe² cot(θ/2)/K

(k = 1/4πε₀ = 9×10⁹ N m² C⁻²)

Atoms

1. Rutherford Scattering Experiment

Observations & Conclusions :

2. Distance of Closest Approach

3. Impact Parameter (b)

4. Bohr's Atomic Model

5. 1st Postulate

Key Points :

6. 2nd Postulate

7. 3rd Postulate

8. Speed (vn) and Radius (rn)

9. Energy of nth Orbit

Kinetic energy :

Electric potential energy :

Total energy : En = Kn + Un

10. Bohr's Model – Summary

11. Hydrogen-like Atoms

12. Hydrogen Atom Energy Levels (Z = 1)

13. Energy Gap (ΔE)

14. Emission & Absorption Spectra

15. Excitation of Electron (by Photons)

If E > 13.6 eV : electron is ejected (free electron)

16. Photons Don't Add Up

17. De-excitation of Electron

18. Number of Spectral Lines

19. Energy & Wavelength of Photon

20. Spectral Series of Hydrogen

21. Other Important Terms (H atom)

22. Other Derived Quantities

Area – graph of ln A vs ln n :

de-Broglie wavelength :

Average force & torque (transition between orbits) :

Key Points :

23. Solved Questions

24. Quick Formula Revision

5. 1^st Postulate

6. 2^nd Postulate

7. 3^rd Postulate

8. Speed (v_n) and Radius (r_n)

9. Energy of n^th Orbit

Total energy : E_n = K_n + U_n