Home › Chemistry › Q
ChemistryAtomic Structure
A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is 4·42 × 10⁻¹⁹ J, how many photons are emitted by the bulb per second ?
Options
1
1·35 × 10¹⁹
2
3·06 × 10¹⁹
3
2·71 × 10¹⁹
4
27·2 × 10¹⁹
Correct Answer
Option 3 : 2·71 × 10¹⁹
Step-by-Step Solution
1

Energy converted to light per second:

Power = 150 W = 150 J/s

Light energy = 8% of 150 = (8/100) × 150 = 12 J/s

2

Number of photons per second:

n = Total light energy / Energy per photon

n = 12 / (4·42 × 10⁻¹⁹)

n = (12 / 4·42) × 10¹⁹

n = 2·714... × 10¹⁹ ≈ 2·71 × 10¹⁹

Light energy/sec = 8% × 150 W = 12 J/s

n = 12 / (4·42×10⁻¹⁹) = 2·71 × 10¹⁹ photons/sec

Theory: Quantum Nature of Light & Atomic Structure
1. Planck's Quantum Theory

Max Planck (1900) proposed that energy is emitted or absorbed in discrete packets called quanta (singular: quantum). For electromagnetic radiation, each quantum is called a photon. Energy of one photon: E = hν = hc/λ, where h = Planck's constant = 6·626 × 10⁻³⁴ J·s, ν = frequency (Hz), c = speed of light = 3 × 10⁸ m/s, λ = wavelength (m). Higher frequency → higher energy per photon. X-rays have much more energy per photon than radio waves. Visible light: ~2–3 eV per photon.

2. Photoelectric Effect — Einstein's Explanation

Einstein (1905, Nobel Prize 1921) explained the photoelectric effect using Planck's quantum concept. When light hits a metal surface, electrons are ejected only if photon energy ≥ work function (Φ) of the metal. KE of ejected electron = hν − Φ. If hν < Φ: no electron ejected regardless of intensity. If hν ≥ Φ: electrons ejected — more intensity means more photons → more electrons, but not faster ones. Threshold frequency ν₀ = Φ/h. Work function Φ = hν₀. This proved light has particle (photon) nature.

3. Energy of a Photon — Key Relations

📌 E = hν (energy-frequency relation)

📌 E = hc/λ (energy-wavelength relation: higher λ → lower E)

📌 E = hcν̃ where ν̃ = wavenumber = 1/λ

📌 1 eV = 1·6 × 10⁻¹⁹ J (electron volt conversion)

📌 Energy of n photons = n × hν

📌 Number of photons n = Total energy / Energy per photon = E_total/hν

📌 In this problem: E per photon = 4·42×10⁻¹⁹ J corresponds to violet/UV light

4. Bohr's Atomic Model — Key Postulates

Bohr (1913) proposed: (1) Electrons revolve in fixed circular orbits (stationary states) without losing energy. (2) Only orbits where mvr = nℏ (n = 1, 2, 3...) are allowed (angular momentum quantisation). (3) Energy is emitted/absorbed only when electron jumps between orbits: ΔE = E₂ − E₁ = hν. For hydrogen: Eₙ = −13·6/n² eV. Radius rₙ = 0·529n² Å (Bohr radius a₀ = 0·529 Å for n=1). Velocity vₙ = 2·18×10⁶/n m/s. Bohr model works only for hydrogen and hydrogen-like ions (He⁺, Li²⁺, etc.).

5. Spectral Lines of Hydrogen

📌 Lyman series: n→1, UV region (1→1 to ∞→1)

📌 Balmer series: n→2, visible region (3→2 to ∞→2) — Hα(red), Hβ(blue-green), Hγ(violet)

📌 Paschen series: n→3, infrared region

📌 Brackett series: n→4, infrared region

📌 Pfund series: n→5, far infrared

📌 Rydberg formula: 1/λ = R_H(1/n₁² − 1/n₂²), R_H = 1·097×10⁷ m⁻¹

6. de Broglie Wavelength — Wave-Particle Duality

de Broglie (1924) proposed that all matter has wave properties. Wavelength λ = h/mv = h/p, where m = mass, v = velocity, p = momentum. For electrons: λ is significant (comparable to atomic dimensions). For macroscopic objects (cricket ball): λ is negligibly small (undetectable). Confirmed by electron diffraction experiments (Davisson-Germer, 1927). For accelerated electron: λ = h/√(2mKE) = h/√(2meV) where V = accelerating voltage. Heavier particles → shorter wavelength (e.g., proton has much shorter λ than electron at same KE).

7. Heisenberg's Uncertainty Principle

It is impossible to simultaneously determine the exact position and exact momentum (or velocity) of a particle: Δx × Δp ≥ h/4π. This is not a limitation of instruments — it's a fundamental property of quantum systems. A particle with definite momentum has completely indefinite position (spread over all space) and vice versa. Consequences: electrons cannot have defined circular orbits (as in Bohr model). Instead, we use probability distributions (orbitals — regions of 90% probability). Uncertainty in energy and time: ΔE × Δt ≥ h/4π.

8. Quantum Numbers — Summary

Principal quantum number n (1, 2, 3...): energy level, shell size. Azimuthal/angular quantum number l (0 to n−1): subshell shape (0=s, 1=p, 2=d, 3=f). Magnetic quantum number m_l (−l to +l): orbital orientation in space. Spin quantum number m_s (+½ or −½): electron spin (up ↑ or down ↓). Pauli exclusion principle: no two electrons in an atom can have all four quantum numbers identical. Maximum electrons in a shell: 2n². Maximum in a subshell: 2(2l+1). Aufbau principle + Hund's rule determine filling order.

Frequently Asked Questions
1. Why do we use only 8% of 150W and not 100%? ⌄
The bulb converts only 8% of electrical energy into light (visible photons). The remaining 92% is wasted as heat (infrared radiation, conducted/convected heat). Only the light energy produces visible photons. So useful light energy per second = 8% × 150 J/s = 12 J/s. The number of photons = 12 J ÷ 4·42×10⁻¹⁹ J/photon = 2·71×10¹⁹ photons/sec. If we mistakenly used 150 J: n = 150/(4·42×10⁻¹⁹) ≈ 3·39×10²⁰ (not an option).
2. What is the energy of a photon of visible light? ⌄
Visible light: λ = 400–700 nm. Energy per photon: E = hc/λ. At λ=400nm (violet): E = (6·626×10⁻³⁴ × 3×10⁸)/(400×10⁻⁹) = 4·97×10⁻¹⁹ J = 3·1 eV. At λ=700nm (red): E = (6·626×10⁻³⁴ × 3×10⁸)/(700×10⁻⁹) = 2·84×10⁻¹⁹ J = 1·77 eV. In this problem: E=4·42×10⁻¹⁹ J ≈ 2·76 eV → corresponds to ~450nm (blue light).
3. How many photons are in 1 mole of light? ⌄
1 mole of photons = 6·022×10²³ photons (Avogadro's number). Energy of 1 mole of photons = Nₐ × hν = Nₐhc/λ. This is called the Einstein (E = Nₐhν). For λ=500nm: E = 6·022×10²³ × (6·626×10⁻³⁴ × 3×10⁸)/(500×10⁻⁹) ≈ 239 kJ/mol. This energy (kJ/mol) is comparable to chemical bond energies — which is why UV light can break chemical bonds (photochemistry, photosynthesis, skin damage).
4. What is the photoelectric effect and which metal properties determine it? ⌄
Photoelectric effect: emission of electrons from a metal surface when light of sufficient frequency hits it. Key observations: (1) Below threshold frequency ν₀: no electron emission regardless of intensity. (2) Above ν₀: electrons emitted instantly — more intense light → more electrons (not faster). (3) KE of electrons = hν − Φ (increases with frequency). Work function Φ = hν₀ depends on the metal: Cs (2·1 eV) — easiest to eject electrons, used in photoelectric cells. Li (2·9 eV), Na (2·75 eV), K (2·3 eV). Pt (6·35 eV) — very hard to eject. Lower Φ metals are used in photosensors.
5. What is the de Broglie wavelength of an electron? ⌄
λ = h/mv. For an electron (m = 9·11×10⁻³¹ kg) moving at 10⁶ m/s: λ = 6·626×10⁻³⁴/(9·11×10⁻³¹ × 10⁶) = 7·27×10⁻¹⁰ m = 0·727 nm. This is comparable to X-ray wavelengths and atomic dimensions! This explains why electron diffraction is possible (electrons diffract off crystal lattices). In contrast, a cricket ball (0·15 kg) at 30 m/s: λ = 6·626×10⁻³⁴/(0·15×30) ≈ 1·5×10⁻³⁴ m — utterly undetectable.
6. What is the energy of an electron in the nth orbit of hydrogen? ⌄
Eₙ = −13·6/n² eV (negative because electron is bound — energy required to free it). n=1 (ground state): E₁ = −13·6 eV. n=2 (first excited state): E₂ = −13·6/4 = −3·4 eV. n=3: E₃ = −13·6/9 = −1·51 eV. n=∞ (ionised): E∞ = 0. Ionisation energy of H from ground state = 13·6 eV. Energy of photon emitted/absorbed = |ΔE| = 13·6(1/n₁² − 1/n₂²) eV. For n₂→n₁ transition (emission): use positive sign. Bohr model gives exact hydrogen spectrum.
7. Why is the Bohr model considered incomplete? ⌄
Bohr model fails for: (1) Multi-electron atoms (He, Li, etc.) — electron-electron repulsions not accounted for. (2) Fine structure of spectral lines (Zeeman effect, Stark effect) — not explained. (3) Violates Heisenberg's uncertainty principle — assumes definite orbit (definite position + momentum simultaneously). (4) Cannot explain bonding, chemical properties. Modern quantum mechanics (Schrödinger equation, orbitals, wave functions) replaced Bohr model and works for all atoms and molecules. Bohr model is still useful for calculating H atom energies and spectra.
8. What is the Rydberg formula and how is it used? ⌄
1/λ = R_H × (1/n₁² − 1/n₂²), where R_H = 1·097×10⁷ m⁻¹ (Rydberg constant), n₁ < n₂. For emission: electron drops from n₂ to n₁, emitting photon. For Balmer series (n₁=2): 1/λ = 1·097×10⁷ × (1/4 − 1/n₂²). For n₂=3 (Hα, red): 1/λ = 1·097×10⁷ × (1/4 − 1/9) = 1·52×10⁶ m⁻¹ → λ = 656 nm (red). For n₂=∞ (series limit): 1/λ = 1·097×10⁷/4 → λ = 364·6 nm (UV). Series limit = minimum wavelength (maximum energy) for that series.
Previous Questions
Q.
Benzene + CH₃Cl (AlCl₃) → toluene; nitration → X + Y; separation method
Organic · Answer: Fractional distillation
Q.
Ti²⁺ (3d²) — calculated spin-only magnetic moment
d-Block · Answer: 2·84 BM
Q.
Pt|H₂(2atm)|HCl(0·02M) — emf using Nernst equation
Electrochemistry · Answer: 0·109 V
Q.
ClF₃ — geometry and lone pairs on Cl atom
p-Block · Answer: T-shaped, 2 lone pairs
Q.
CH₃CH₂CH₂OH + PCl₅ → X; alc.KOH → Y; HBr/peroxide → Z
Haloalkanes · Answer: X=POCl₃, Z=1-bromopropane