How to Use This Page
This page is designed to help you prepare systematically for the CBSE Class 12 Physics board exam (Part 2 chapters). Each section mirrors the question formats you will face — MCQs, fill-in-the-blanks, short answers, and detailed numerical/derivation problems. Work through them in order: attempt first, then compare with the model answers.
Start by revising the Key Concepts section to refresh formulas and definitions. Then tackle the MCQs and short answers to test recall. Finally, work through the Answer in Detail section — these carry the most marks and require structured, step-by-step responses with proper diagrams and units.
Do not memorize answers word for word. Instead, internalize the method: identify the concept, write the relevant formula, substitute values with correct units, and state the conclusion clearly. The board exam rewards clarity, correct units, and neat diagrams more than lengthy prose.
Allocate your exam time wisely: MCQs and short answers should take 25-30 minutes total, leaving 90-100 minutes for the 5-mark detailed answers and numerical problems. Always draw a labelled diagram when a question involves an optical device, interference pattern, or circuit.
Key Concepts (Quick Recap)
Chapter 9: Ray Optics
- Laws of reflection: angle of incidence = angle of reflection
- Snell's law: n₁ sin i = n₂ sin r
- Lens maker's equation: 1/f = (n - 1)(1/R₁ - 1/R₂)
- Power of lens: P = 1/f (in metres), unit: dioptre (D)
- Prism formula: n = sin[(A + Dₘ)/2] / sin(A/2), where Dₘ = minimum deviation
- Mirror formula: 1/v + 1/u = 1/f
- Magnification: m = -v/u (mirror), m = v/u (lens)
- Angular magnification of simple microscope: M = 1 + D/f (D = 25 cm, least distance of distinct vision)
- Angular magnification of compound microscope: M = (vₒ/uₒ)(D/fₑ)
- Angular magnification of refracting telescope: M = -fₒ/fₑ
- Total internal reflection: sin i_c = n₂/n₁ (n₁ > n₂)
- Combination of thin lenses in contact: 1/f = 1/f₁ + 1/f₂ + ...
Chapter 10: Wave Optics
- Huygens' principle: every point on a wavefront acts as a source of secondary wavelets
- Young's double slit experiment: fringe width β = λD/d
- Condition for constructive interference: path difference = nλ
- Condition for destructive interference: path difference = (2n - 1)λ/2
- Intensity: I = I₁ + I₂ + 2√(I₁I₂) cos φ; for equal intensities, I_max = 4I₀, I_min = 0
- Diffraction at single slit: a sin θ = nλ (minima), a sin θ = (2n + 1)λ/2 (secondary maxima)
- Resolving power of telescope: R.P. = D/1.22λ
- Resolving power of microscope: R.P. = 2μ sin θ/λ = 2NA/λ
- Brewster's law: tan i_B = n₂/n₁ = μ
- Malus's law: I = I₀ cos²θ
- Degree of polarization: P = (I_max - I_min)/(I_max + I_min)
Chapter 11: Dual Nature of Radiation and Matter
- Photoelectric effect: Einstein's equation K_max = hν - φ₀ (where φ₀ = work function)
- Threshold frequency: ν₀ = φ₀/h
- Stopping potential: eV₀ = hν - φ₀
- de Broglie wavelength: λ = h/p = h/mv = h/√(2mE)
- For electron accelerated through V volts: λ = 1.227/√V nm (≈ 1.227/√V Å)
- Davisson-Germer experiment confirmed wave nature of electrons
- Photoelectric effect depends on frequency, not intensity (above threshold)
- Number of photoelectrons ∝ intensity of light
Chapter 12: Atoms
- Bohr's postulates: (1) stationary orbits with quantised angular momentum mvr = nh/2π, (2) radiation emitted/absorbed when electron transitions between orbits with ΔE = hν
- Bohr radius: a₀ = 0.529 Å = 0.529 × 10⁻¹⁰ m
- Energy of hydrogen electron: E_n = -13.6/n² eV
- Radius of nth orbit: r_n = n²a₀ = n² × 0.529 Å
- Velocity in nth orbit: v_n = v₁/n = 2.19 × 10⁶/n m/s
- Rydberg formula: 1/λ = R(1/n_f² - 1/n_i²), R = 1.097 × 10⁷ m⁻¹
- Lyman series: n_f = 1 (UV); Balmer: n_f = 2 (visible); Paschen: n_f = 3 (IR); Brackett: n_f = 4; Pfund: n_f = 5
- Impact parameter: b = Ze² cot(θ/2) / (4πε₀ × ½mv²)
Chapter 13: Nuclei
- Nuclear radius: R = R₀A^(1/3), R₀ = 1.2 fm = 1.2 × 10⁻¹⁵ m
- Mass-energy equivalence: E = Δmc² (1 amu = 931.5 MeV)
- Binding energy: B.E. = (Zm_p + Nm_n - M_nucleus)c²
- B.E. per nucleon peaks at Fe-56 (~8.75 MeV), indicating greatest stability
- Radioactivity law: N = N₀e^(-λt), where λ = decay constant
- Half-life: T₁/₂ = 0.693/λ = ln 2/λ
- Activity: A = λN = A₀e^(-λt), unit: becquerel (Bq) = 1 disintegration/s
- Alpha decay: energy shared as kinetic energy; Q = (M_X - M_Y - M_He)c²
- Beta decay: n → p + e⁻ + ν̄ (β⁻); p → n + e⁺ + ν (β⁺)
- Nuclear fission: heavy nucleus splits; energy ~200 MeV per fission of U-235
- Nuclear fusion: light nuclei combine; energy per unit mass much higher than fission
Chapter 14: Semiconductor Electronics
- Intrinsic semiconductor: n_e = n_h = n_i; n_i² = n_e × n_h
- Extrinsic: n-type (donor doped, n_e >> n_h); p-type (acceptor doped, n_h >> n_e)
- n_e × n_h = n_i² (mass action law)
- p-n junction: depletion width ~0.1 μm, barrier potential ~0.7 V (Si), ~0.3 V (Ge)
- Forward bias: depletion width decreases, current flows
- Reverse bias: depletion width increases, very small leakage current
- Zener diode: operates in reverse breakdown; used as voltage regulator
- Transistor: three regions — emitter (heavily doped), base (thin, lightly doped), collector (moderately doped)
- Transistor current relation: I_E = I_B + I_C; α = I_C/I_E; β = I_C/I_B; β = α/(1-α)
- Common emitter current gain: β = ΔI_C/ΔI_B (typically 20-500)
- Logic gates: AND (Y = A·B), OR (Y = A+B), NOT (Y = Ā), NAND (Y = ∼(A·B)), NOR (Y = ∼(A+B))
- Truth tables for all basic gates and combinations
I. Choose the Best Answer (MCQs)
1. A ray of light travelling in water (n = 4/3) is incident on a glass plate (n = 3/2) at an angle of 30°. The angle of refraction inside the glass is:
Answer
(a) sin⁻¹(4/3√3) (b) sin⁻¹(4/9) ✅ (c) sin⁻¹(2/3) (d) 30°
ExplanationUsing Snell's law: n_w sin i = n_g sin r → (4/3) sin 30° = (3/2) sin r → (4/3)(1/2) = (3/2) sin r → sin r = (4/3)(1/2)(2/3) = 4/9 → r = sin⁻¹(4/9).
2. In Young's double slit experiment, if the distance between slits is halved and the screen distance is doubled, the fringe width becomes:
Answer
(a) 4 times ✅ (b) 2 times (c) halved (d) remains same
Explanationβ = λD/d. If d → d/2 and D → 2D, then β' = λ(2D)/(d/2) = 4λD/d = 4β.
3. The stopping potential for a metal surface is V₀ when the incident light has frequency ν. If the frequency is doubled, the stopping potential becomes:
Answer
(a) 2V₀ (b) V₀ + hν/e ✅ (c) 2V₀ + hν/e (d) V₀ + hν/2e
ExplanationeV₀ = hν - φ₀. If frequency becomes 2ν: eV₀' = h(2ν) - φ₀ = 2hν - φ₀ = (hν - φ₀) + hν = eV₀ + hν → V₀' = V₀ + hν/e.
4. The ratio of radii of the first three Bohr orbits in a hydrogen atom is:
Answer
(a) 1 : 2 : 3 (b) 1 : 4 : 9 ✅ (c) 1 : 1/2 : 1/3 (d) 1 : 1/4 : 1/9
Explanationr_n = n²a₀, so r₁ : r₂ : r₃ = 1² : 2² : 3² = 1 : 4 : 9.
5. The binding energy per nucleon for a medium mass nucleus is approximately:
Answer
(a) 200 MeV (b) 8.5 MeV ✅ (c) 2.8 MeV (d) 0.8 MeV
ExplanationFor medium nuclei (A ≈ 56, Fe), B.E. per nucleon is about 8.5 MeV — the highest among all nuclei, explaining why iron is the most stable.
6. In a p-n junction diode, the depletion region is formed due to:
Answer
(a) Diffusion of majority carriers ✅ (b) Drift of minority carriers (c) Both diffusion and drift equally (d) External bias voltage
ExplanationMajority carriers (electrons from n-side, holes from p-side) diffuse across the junction and recombine, leaving behind immobile ions that create the depletion region with a built-in potential.
7. A transistor has α = 0.98. The value of β is:
Answer
(a) 98 (b) 4.9 ✅ (c) 49 (d) 0.49
Explanationβ = α/(1-α) = 0.98/(1-0.98) = 0.98/0.02 = 49.
8. The Boolean expression Y = ∼(A · B) represents:
Answer
(a) AND gate (b) OR gate (c) NAND gate ✅ (d) NOR gate
ExplanationY = ∼(A · B) = NOT(A AND B) = NAND gate output.
9. Two coherent sources of intensity I₀ and 4I₀ interfere. The ratio of maximum to minimum intensity is:
Answer
(a) 9:1 ✅ (b) 3:1 (c) 4:1 (d) 2:1
ExplanationI_max = (√I₁ + √I₂)² = (√I₀ + √4I₀)² = (3√I₀)² = 9I₀; I_min = (√I₁ - √I₂)² = (√I₀ - 2√I₀)² = I₀. Ratio = 9:1.
10. The energy of an electron in the second Bohr orbit of hydrogen atom is:
Answer
(a) -13.6 eV (b) -3.4 eV ✅ (c) -1.51 eV (d) -0.85 eV
ExplanationE_n = -13.6/n² eV. For n = 2: E₂ = -13.6/4 = -3.4 eV.
II. Fill in the Blanks
1. The focal length of a lens is +20 cm. Its power is +5 D (P = 1/f in metres = 1/0.20 = 5 D; positive f means converging lens).
2. In Young's double slit experiment, if white light is used instead of monochromatic light, the central fringe is white and the fringes are coloured with blue closer to the central fringe and red farther away.
3. The work function of a metal is 2 eV. The threshold wavelength is 6200 Å (or 620 nm). [λ₀ = hc/φ₀ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(2 × 1.6 × 10⁻¹⁹) ≈ 6.2 × 10⁻⁷ m = 620 nm]
4. According to Bohr's model, the angular momentum of the electron in the nth orbit is nh/2π (or nh/2π, where n = 1, 2, 3, ...).
5. The nuclear radius of ²⁷Al is approximately 3.6 fm. [R = R₀A^(1/3) = 1.2 × (27)^(1/3) = 1.2 × 3 = 3.6 fm]
6. The mass action law in semiconductors states that n_e × n_h = n_i².
7. In a common-emitter transistor amplifier, the phase difference between input and output signals is 180° (π radians).
8. The phenomenon of total internal reflection occurs when light travels from a denser to rarer medium and the angle of incidence exceeds the critical angle.
9. Brewster's angle for light incident from air (n = 1) on glass (n = 1.5) is tan⁻¹(1.5) ≈ 56.3°.
10. The half-life of a radioactive substance is 10 years. After 30 years, the fraction of the substance that remains is 1/8. [N = N₀(1/2)^(t/T) = N₀(1/2)^(30/10) = N₀/8]
III. True or False
1. False. The focal length of a convex lens increases when it is immersed in water (n_water = 4/3). Correction: The focal length increases because the relative refractive index (n_g/n_w) decreases, reducing the lens power. Since n_glass/n_water < n_glass/n_air, f increases.
2. True. In interference, the fringes are of equal width, unlike in single-slit diffraction where central maxima is wider than secondary maxima.
3. False. The photoelectric effect proves that light has a particle nature, not a wave nature. Correction: The photoelectric effect demonstrates the particle (photon) nature of light. Phenomena like interference and diffraction demonstrate the wave nature.
4. True. According to Bohr's model, the velocity of an electron in a hydrogen atom is inversely proportional to the principal quantum number (v ∝ 1/n).
5. False. The binding energy per nucleon increases with mass number for light nuclei and decreases for heavy nuclei. Correction: B.E. per nucleon first increases (for A < 56), reaches a maximum around iron (A ≈ 56), and then decreases for heavier nuclei.
6. True. A Zener diode is operated in the reverse breakdown region and is used as a voltage regulator.
7. False. An AND gate can be made using two NAND gates. Correction: An AND gate can indeed be made using two NAND gates (first NAND gives NAND output, second NAND with both inputs tied together acts as NOT, giving AND). This statement is actually true — the original claim that it is false was wrong. AND gate = NAND followed by NOT (which is another NAND with shorted inputs).
8. True. The energy of a photon is directly proportional to its frequency (E = hν) and inversely proportional to its wavelength (E = hc/λ).
IV. Match the Following
| Column A | Column B |
|---|
| 1. Snell's law | (e) n₁ sin i = n₂ sin r |
| 2. Malus's law | (f) I = I₀ cos²θ |
| 3. Einstein's photoelectric equation | (a) K_max = hν - φ₀ |
| 4. Bohr's quantisation condition | (c) mvr = nh/2π |
| 5. Mass-energy equivalence | (b) E = Δmc² |
| 6. Mass action law | (d) n_e × n_h = n_i² |
| 7. Fringe width (YDSE) | (h) β = λD/d |
| 8. Half-life | (g) T₁/₂ = 0.693/λ |
V. Answer Briefly
1. State the laws of reflection and refraction.
Answer
Law of Reflection: (i) The angle of incidence equals the angle of reflection (i = r). (ii) The incident ray, reflected ray, and the normal at the point of incidence all lie in the same plane. Law of Refraction (Snell's Law): (i) The incident ray, refracted ray, and the normal at the point of incidence all lie in the same plane. (ii) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media: n₁ sin i = n₂ sin r.
2. What is total internal reflection? State the conditions for it to occur.
Answer
Total internal reflection is the phenomenon in which a ray of light travelling from a denser medium to a rarer medium is completely reflected back into the denser medium when the angle of incidence exceeds the critical angle. Conditions: (i) Light must travel from a denser to a rarer medium (n₁ > n₂). (ii) The angle of incidence must be greater than the critical angle i_c, where sin i_c = n₂/n₁. Applications include optical fibres, mirages, diamond brilliance, and totally reflecting prisms in binoculars.
3. Distinguish between interference and diffraction.
Answer
| Feature | Interference | Diffraction | |---|---|---| | Source | Two coherent sources | Single slit/aperture | | Fringe width | All fringes equal width | Central fringe wider | | Intensity | All bright fringes same intensity | Intensity decreases for higher orders | | Superposition | Waves from two sources | Wavelets from parts of same wavefront |
4. What is the photoelectric effect? Why can it not be explained by the wave theory of light?
Answer
The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. Wave theory fails because: (i) It cannot explain why photoelectric emission is instantaneous (no time lag), even for very low intensity. (ii) It cannot explain the existence of a threshold frequency below which no emission occurs regardless of intensity. (iii) It cannot explain why K_max depends on frequency and not on intensity. (iv) Wave theory predicts that increasing intensity should increase K_max, which is experimentally false.
5. State Bohr's postulates for the hydrogen atom.
Answer
Postulate 1 (Stationary orbits): An electron revolves in certain stable orbits (stationary orbits) without radiating energy. In these orbits, the angular momentum is quantised: mvr = nh/2π, where n = 1, 2, 3, ... Postulate 2 (Radiation condition): Radiation of frequency ν is emitted or absorbed when the electron jumps between orbits of energies E_i and E_f: hν = E_i - E_f. If E_i > E_f, radiation is emitted; if E_i < E_f, radiation is absorbed.
6. Define binding energy and binding energy per nucleon. What is their significance?
Answer
Binding energy of a nucleus is the energy required to separate all its nucleons (protons and neutrons) completely from each other. It equals the mass defect multiplied by c²: B.E. = Δm × c² = (Zm_p + Nm_n - M_nucleus)c². Binding energy per nucleon = B.E./A. It indicates nuclear stability — higher B.E. per nucleon means greater stability. Iron-56 has the highest B.E. per nucleon (~8.75 MeV), making it the most stable nucleus.
7. Distinguish between n-type and p-type semiconductors.
Answer
| Feature | n-type | p-type | |---|---|---| | Dopant | Pentavalent (As, P, Sb) | Trivalent (B, In, Al) | | Majority carriers | Electrons (n_e >> n_h) | Holes (n_h >> n_e) | | Minority carriers | Holes | Electrons | | Donor/Acceptor level | Donor level near CB | Acceptor level near VB | | Conductivity | σ = n_e e μ_e | σ = n_h e μ_h |
8. What is a Zener diode? How does it regulate voltage?
Answer
A Zener diode is a specially designed p-n junction diode that operates in the reverse breakdown region. It is heavily doped, so the depletion region is very thin and breakdown occurs at a precise voltage (V_Z). Voltage regulation: The Zener diode is connected in reverse bias across the load. When the input voltage increases, the current through the Zener increases while the voltage across it (and hence across the load) remains nearly constant at V_Z. The series resistor limits the current. Any change in input voltage is absorbed by a change in Zener current, keeping the output voltage stable.
VI. Answer in Detail
1. Derive the lens maker's equation for a thin convex lens. A convex lens made of glass (n = 1.5) has surfaces with radii of curvature +20 cm and -30 cm. Find its focal length and power.
Answer
Derivation: Consider a thin lens of refractive index n_l in a medium of refractive index n_m. Light from object O passes through the first surface (radius R₁), forming an intermediate image I' at distance v'. Using the single-surface refraction formula: n_l/v' - n_m/u = (n_l - n_m)/R₁ ... (1) This intermediate image I' acts as a virtual object for the second surface (radius R₂). The final image I is formed at distance v: n_m/v - n_l/v' = (n_m - n_l)/R₂ ... (2) Adding equations (1) and (2): n_m/v - n_m/u = (n_l - n_m)(1/R₁ - 1/R₂) Dividing by n_m: 1/v - 1/u = (n_l/n_m - 1)(1/R₁ - 1/R₂) For a lens in air, n_m = 1 and n_l = n: 1/v - 1/u = (n - 1)(1/R₁ - 1/R₂) ... (Lens maker's equation) For parallel rays (u → ∞), v = f: 1/f = (n - 1)(1/R₁ - 1/R₂) Numerical Problem: Given: n = 1.5, R₁ = +20 cm, R₂ = -30 cm 1/f = (1.5 - 1)(1/20 - 1/(-30)) 1/f = 0.5 × (1/20 + 1/30) 1/f = 0.5 × (3 + 2)/60 = 0.5 × 5/60 1/f = 0.5 × 1/12 = 1/24 f = +24 cm = 0.24 m Power P = 1/f = 1/0.24 = +4.17 D The positive focal length and power confirm this is a converging (convex) lens.
2. Derive the expression for fringe width in Young's double slit experiment. Two slits 0.5 mm apart are illuminated by light of wavelength 6000 Å. The screen is 1.5 m away. Calculate the fringe width and the distance of the 4th bright fringe from the central fringe.
Answer
Derivation: Let two coherent sources S₁ and S₂ be separated by distance d. Screen is at distance D from the slits (D >> d). Consider a point P on the screen at distance y from the central maximum O. Path difference at P = S₂P - S₁P = d sin θ ≈ d × y/D (for small θ) For bright fringes (constructive interference): Path difference = nλ (n = 0, 1, 2, ...) d × y_n/D = nλ y_n = nλD/d (position of nth bright fringe) Fringe width = distance between consecutive bright fringes: β = y_{n+1} - y_n = (n+1)λD/d - nλD/d = λD/d For dark fringes: Path difference = (2n - 1)λ/2 → y_n = (2n - 1)λD/(2d) Numerical Problem: Given: d = 0.5 mm = 0.5 × 10⁻³ m, λ = 6000 Å = 6 × 10⁻⁷ m, D = 1.5 m Fringe width β = λD/d = (6 × 10⁻⁷ × 1.5)/(0.5 × 10⁻³) β = (9 × 10⁻⁷)/(5 × 10⁻⁴) = 1.8 × 10⁻³ m = 1.8 mm Position of 4th bright fringe (n = 4): y₄ = 4λD/d = 4 × 1.8 mm = 7.2 mm from the centre.
3. State and explain Einstein's photoelectric equation. Light of wavelength 4000 Å is incident on a metal surface with work function 2 eV. Calculate: (a) the maximum kinetic energy of photoelectrons, (b) the stopping potential, and (c) the threshold wavelength.
Answer
Einstein's Photoelectric Equation: Einstein explained the photoelectric effect using the photon concept. When a photon of energy hν strikes a metal surface, its energy is used in two ways: (i) A minimum energy φ₀ (work function) is needed to overcome the surface barrier and free the electron from the metal. (ii) The remaining energy appears as the kinetic energy of the emitted photoelectron. K_max = hν - φ₀ This is Einstein's photoelectric equation. Key implications: - If hν < φ₀ (i.e., ν < ν₀), no emission occurs regardless of intensity. - K_max depends on frequency, not intensity. - Number of photoelectrons ∝ intensity (more photons = more electrons). - Emission is instantaneous (no time lag). Stopping potential V₀: K_max = eV₀, so eV₀ = hν - φ₀ Numerical Problem: Given: λ = 4000 Å = 4 × 10⁻⁷ m, φ₀ = 2 eV (a) Energy of photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J Converting to eV: E = 4.97 × 10⁻¹⁹/(1.6 × 10⁻¹⁹) = 3.1 eV K_max = hν - φ₀ = 3.1 - 2 = 1.1 eV (b) Stopping potential: eV₀ = K_max = 1.1 eV → V₀ = 1.1 V (c) Threshold wavelength: φ₀ = hc/λ₀ → λ₀ = hc/φ₀ λ₀ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(2 × 1.6 × 10⁻¹⁹) = 6.22 × 10⁻⁷ m = 6220 Å ≈ 622 nm Since the incident wavelength (400 nm) is less than the threshold wavelength (622 nm), photoelectric emission will occur.
4. Derive the expression for the radius and energy of the nth Bohr orbit. Calculate the first three energy levels of hydrogen and identify the spectral series for the transition from n = 3 to n = 2.
Answer
Derivation of radius: From Bohr's quantisation condition: m v_n r_n = nh/2π ... (1) For a stable orbit, centripetal force = Coulomb force: mv_n²/r_n = (1/4πε₀)(e²/r_n²) ... (2) From (1): v_n = nh/(2πmr_n). Substituting in (2): m × n²h²/(4π²m²r_n²) = e²/(4πε₀r_n) r_n = n²h²ε₀/(πme²) = n² × a₀ where a₀ = h²ε₀/(πme²) = 0.529 Å (Bohr radius) r_n = n² × 0.529 Å Derivation of energy: Total energy = K.E. + P.E. = ½mv_n² + (-e²/4πε₀r_n) From (2): ½mv_n² = e²/(8πε₀r_n) E_n = e²/(8πε₀r_n) - e²/(4πε₀r_n) = -e²/(8πε₀r_n) Substituting r_n: E_n = -(me⁴)/(8ε₀²n²h²) = -13.6/n² eV E_n = -13.6/n² eV Numerical values: E₁ = -13.6/1 = -13.6 eV E₂ = -13.6/4 = -3.4 eV E₃ = -13.6/9 = -1.51 eV For transition n = 3 → n = 2: ΔE = E₃ - E₂ = -1.51 - (-3.4) = 1.89 eV This corresponds to the Balmer series (visible region, n_f = 2). Wavelength: λ = hc/ΔE = 1240/1.89 ≈ 656 nm (Hα line, red).
5. Define half-life and derive the relation between half-life and decay constant. A radioactive sample has a half-life of 5 years. What fraction of the sample remains after 15 years? Also calculate the activity if the initial number of atoms is 10²⁰.
Answer
Definition: Half-life (T₁/₂) is the time in which half the radioactive atoms present in a sample disintegrate. Derivation: From the radioactive decay law: N = N₀ e^(-λt) At t = T₁/₂: N = N₀/2 N₀/2 = N₀ e^(-λT₁/₂) 1/2 = e^(-λT₁/₂) Taking natural log: ln(1/2) = -λT₁/₂ -ln 2 = -λT₁/₂ T₁/₂ = ln 2/λ = 0.693/λ Numerical Problem: Given: T₁/₂ = 5 years, t = 15 years, N₀ = 10²⁰ Number of half-lives = t/T₁/₂ = 15/5 = 3 Fraction remaining = (1/2)³ = 1/8 = 0.125 = 12.5% Alternative method: λ = 0.693/T₁/₂ = 0.693/5 = 0.1386 per year N = N₀ e^(-λt) = 10²⁰ × e^(-0.1386 × 15) = 10²⁰ × e^(-2.079) ≈ 10²⁰ × 0.125 = 1.25 × 10¹⁹ Activity: A = λN = 0.1386 × 1.25 × 10¹⁹ = 1.73 × 10¹⁸ per year Converting to Bq: 1 year = 3.156 × 10⁷ s λ = 0.1386/(3.156 × 10⁷) = 4.39 × 10⁻⁹ s⁻¹ A = λN = 4.39 × 10⁻⁹ × 1.25 × 10¹⁹ = 5.49 × 10¹⁰ Bq ≈ 5.49 × 10¹⁰ disintegrations/s
6. Draw the V-I characteristics of a p-n junction diode in forward and reverse bias. Explain how a diode can be used as a half-wave rectifier.
Answer
V-I Characteristics: Forward Bias: Initially, current is very small until the barrier potential is overcome (≈0.7 V for Si, ≈0.3 V for Ge). Beyond the knee voltage, current rises rapidly with small increase in voltage. The dynamic resistance r_d = ΔV/ΔI is very small (a few ohms). Reverse Bias: A very small leakage current (μA range) flows due to minority carriers. At the breakdown voltage (V_br), current increases sharply. For a Zener diode, this breakdown is non-destructive and occurs at a well-defined voltage. Half-Wave Rectifier: Circuit: AC source → diode (forward biased during positive half) → load resistor R_L. Working: During the positive half-cycle of AC input, the diode is forward biased and conducts. Current flows through R_L, producing output voltage. During the negative half-cycle, the diode is reverse biased and does not conduct. No current flows through R_L, so output is zero. Result: Only the positive half-cycles appear across the load. The output is pulsating DC with frequency equal to the input frequency. Efficiency: η = 0.406 = 40.6% (theoretical maximum for half-wave rectifier) Ripple factor: γ = 1.21 Limitation: Half-wave rectifier wastes the negative half-cycle. A full-wave rectifier (using two diodes and a centre-tapped transformer, or four diodes in a bridge configuration) is preferred for higher efficiency (η = 81.2%) and lower ripple factor (γ = 0.48).
Common Mistakes to Avoid
1. Sign convention errors in ray optics: Always use the New Cartesian Sign Convention consistently. Distances measured in the direction of incident light are positive; against it are negative. Lens maker's equation uses R₁ and R₂ with proper signs — a common source of error in numerical problems.
2. Confusing fringe width with angular fringe width: Fringe width β = λD/d (linear, on screen) while angular fringe width θ = λ/d. These are different quantities with different units. Read the question carefully to determine which is asked.
3. Using wrong formula for de Broglie wavelength: Remember λ = h/√(2mE) where E is kinetic energy. For an electron accelerated through potential V, use λ = 1.227/√V nm. Do not mix up V (voltage) with v (velocity).
4. Incorrect energy level calculations in Bohr model: The energy E_n = -13.6/n² eV gives the total energy (negative). The kinetic energy is +13.6/n² eV and the potential energy is -27.2/n² eV. Remember that K.E. = -E_n and P.E. = 2E_n. Do not use positive 13.6 for total energy.
5. Confusing fission with fusion energy calculations: In fission, a heavy nucleus (like U-235) splits into lighter nuclei — energy per fission ≈ 200 MeV. In fusion, light nuclei (like H isotopes) combine — energy per fusion ≈ 17.6 MeV but energy per unit mass is much higher. Do not confuse the two processes.
6. Sign errors in semiconductor currents: In a transistor, I_E = I_B + I_C always. Current gain α = I_C/I_E < 1, while β = I_C/I_B >> 1. The relation β = α/(1-α) gives β > 1 since α < 1. In common-emitter configuration, the output is 180° out of phase with the input.
Exam Checklist
- [ ] Can you write the lens maker's equation with correct sign convention and solve numerical problems on focal length and power?
- [ ] Can you derive fringe width (β = λD/d) and solve YDSE numerical problems including conditions for bright/dark fringes?
- [ ] Do you know Einstein's photoelectric equation, the relation between stopping potential and frequency, and can you calculate threshold wavelength and K_max?
- [ ] Can you state Bohr's postulates, derive expressions for r_n and E_n, and identify spectral series (Lyman, Balmer, Paschen)?
- [ ] Can you derive the half-life relation T₁/₂ = 0.693/λ and solve numerical problems on radioactive decay (fraction remaining, activity)?
- [ ] Can you explain p-n junction formation, forward/reverse bias characteristics, and working of Zener diode as voltage regulator?
- [ ] Can you draw and explain truth tables for all basic logic gates (AND, OR, NOT, NAND, NOR) and their combinations?
- [ ] Do you carry correct units throughout numerical problems and write the final answer with proper units (eV, nm, Bq, D, etc.)?
