### SECTION A
**Answer ALL the questions in this section**
1. State one fundamental similarity and difference between diffuse and regular scattering of a light wave. (ray diagram is not needed).
2. Two lenses (converging lens of focal length 30 cm) and the other is (diverging lens of focal length 40 cm) are placed in contact to each other. Calculate the focal length and power of the combination.
3. Determine the approximate number of electrons passing through a wire per minute if the current is 1.00 mA.
4. Two equal resistors of resistances xΩ each, are connected in parallel, the resultant combination is connected in series with another resistor of resistances xΩ. Calculate the effective resistance of the combination.
5. What is a galvanometer? If a galvanometer is connected (i) in series (ii) in parallel to a circuit. State in each case the possible quantity to be measured by the galvanometer.
6. A uniform electric field is obtained between two parallel plates separated by 20mm apart. if the potential difference applied is 10 V which cause the electrons initially at rest to move by the field, calculate the intensity of the field and force acting on the electrons.
7. Consider an electron moving with a speed of 1.5×10⁷ms⁻¹ perpendicular to a magnetic field of uniform flux density of 0.0012T. Calculate the force on the electron.
8. Radiation with a wavelength of 200 nm strikes a metal surface in a vacuum. The ejected electrons have a maximum speed of 7.22 ×10⁵ms⁻¹. What is the work function of the metal?
9. The element ¹⁵₃₂P decays by β and γ-emission to a nucleus ᴬ_Z X. Determine are the values of A and Z?
10. A radioactive source has a half-life of 23 days. Find the decay constant.
---
### SECTION B: GEOMETRIC OPTICS
**Answer ONE (1) question only from this section.**
11. a. For each of the following questions i, ii and iii, support your answer with labeled ray diagrams:
- i. distinguish between ray and beams of monochromatic light.
- ii. state the laws of reflection of red light.
- iii. distinguish between regular and diffuse reflection of red light.
11(b). Suppose an object is placed at the centre of curvature of a concave mirror. Find the image distance and the magnification produced. Sketch its ray diagram and comment on your answer.
c. An object 27 cm is placed in front of a concave mirror of focal length 18 cm, at what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained. Find the size and nature of the image.
---
12. a.
- i. A ray of light is incident on one face of a triangular glass prism, after refraction it emerges on the other face. Sketch the ray diagram and indicate the following: **angle of emergence, angle of deviation, angle of prism and angle of refraction.**
- ii. hence, explain the underlined term.
- iii. state the condition for each angle of minimum and maximum deviations to occur.
bi. Show that for a small angled prism (A), the expression:
μ = sin((A + δₘᵢₙ)/2) / sin(A/2)
reduces to δₘᵢₙ = (μ−1)A. Show the sketch of angle of deviation against angle of incidence.
- ii. when white light passes through a 6° prism, the refractive indices for red and violet colours are 1.641 and 1.659 respectively. What is the angular dispersion produced by the prism?
- iii. Consider an equilateral prism having the refractive index 1.5 of its material. Calculate the angle of incidence of a ray of light for the minimum deviation.
---
### SECTION C: ELECTRICITY AND MAGNETISM
**Answer any TWO (2) questions from this section.**
**13.a.**
- (i) Briefly explain the concept of charging by friction. Give an example.
- (ii) Consider three isolated electric charges A, B, and C. when they are brought close to each other, the electric charges A and B attract each other and B and C repel each other. Determine the possible charges of A, B, and C. What would happened when A and C are brought close together?
(b). A glass rod is rubbed with a silk cloth. The glass rod acquires a charge of +19.2 × 10⁻¹⁹ C.
- (i) What is the magnitude of negative charge acquired by the silk?
- (ii) Determine the number of electrons lost by glass rod.
- (iii) Compute the force of attraction between sodium and chlorine ions in salt molecules if each ion carries one electronic charge and the separation is 2.8 × 10⁻¹⁰ m?
**14a.**
- (i) Briefly explain the terms conductance (G) and conductivity (σ) as applied to electricity.
- (ii) Express the conductance in terms of conductivity.
b. A wire of length 8 m and cross-sectional area 3 mm² has a resistance of 0.16 Ω. If the same wire of the same length with cross-sectional area is 1 mm², determine the resistance of the wire.
c. A flux of 25 mWb links with a 1500 turn coil when a current of 3 A passes through the coil. Calculate the inductance of the coil and the energy stored in the magnetic field.
**15a.**
- i. Write down the integral form of the expression of the Ampere's law that describes the magnetic field produced by a current-carrying wire. Use it to find the expression for the magnetic field strength.
- b. Calculate the magnetic field strength 7.0 m below a power line carrying 12,000 A of current.
c. A 2μF capacitor is charged by 12V, D.C supply. Wires connecting the capacitor to the battery are then disconnected from the battery and connected in parallel with a 4μF capacitor. Calculate the charge stored in the 2μF capacitor initially and finally after connecting the 4μF capacitor.
**16a.**
- (i) State three differences between alternating and direct current.
- (ii) A capacitor of capacitance C is being charged through a resistor of resistance R. Show that the quantity RC has the unit of time.
b. From figure 1, find the effective capacitance between the terminal **ab.**
*(Figure 1 shows: 50μF and 60μF in series on top branch; 70F and 20μF and 120μF in bottom branch, connected between terminals a and b with Ceq)*
c. A coil having a resistance of 10Ω and an inductance of 125 mH is connected in series with a 60μF capacitor across a 120V supply. At what frequency does resonance occur? Find the current flowing at the resonant frequency. What is the Q-factor of this circuit at resonance?
---
### SECTION D: MODERN PHYSICS
**Answer ONE (1) question from this section.**
**17a.**
- i. Explain the concept of photoelectric effect that leads to the Einstein's photoelectric equation.
- ii. Obtain the Einstein's photoelectric equation in terms of cutoffs wavelength. Sketch its graph. From the graph, indicate clearly the points of work function and threshold wave length. From the slope of the graph, find Planck's constant.
b. In a photoelectric effect, it was found that for light of wavelength 4000 Å, a stopping potential of 2V is needed and for light of wavelength 6000 Å, a stopping potential of 1V is needed. Use these data to deduce the work function of the material and the Planck's constant.
c. The energy in Bohr nth orbit is given by −13.6/n² eV. Calculate the energy when the electron jumps from the third state to the second.
**18a.** Define the terms half-life and decay constant of a radioactive material. Deduce the relation between the two quantities.
---
b. In the decay of radium-226, an alpha particle of energy 4.99eV is released. Calculate the energy released and show that, this is consisted with the principle of conservation of mass-energy relation. (²²⁶₈₈Ra = 226.0245u, ²²²₈₆Rn = 222.0175u, ⁴₂He = 4.0026u, 1u = 1.66×10⁻²⁷ kg).
18c. Consider a proton in the nucleus. If its uncertainty in position is 10⁻¹⁴ m. What is the uncertainty in momentum of the electron? Calculate its kinetic energy.
---
# COMPLETE SOLUTIONS
## SECTION A
### Q1. Similarity and Difference Between Diffuse and Regular Scattering
**Similarity:**
Both diffuse and regular (specular) reflection obey the **laws of reflection** — the angle of incidence equals the angle of reflection at each point of the surface.
**Difference:**
- **Regular reflection:** Occurs on **smooth/polished surfaces**; parallel incident rays reflect as parallel rays, producing a clear image.
- **Diffuse reflection:** Occurs on **rough/uneven surfaces**; parallel incident rays reflect in **many different directions**, producing no clear image.
---
### Q2. Two Lenses in Contact — Focal Length and Power
**Given:**
- Converging lens: f₁ = +30 cm
- Diverging lens: f₂ = −40 cm
**Formula for lenses in contact:**
1/f = 1/f₁ + 1/f₂
1/f = 1/30 + 1/(−40)
1/f = 4/120 − 3/120
1/f = 1/120
**f = 120 cm = 1.2 m**
**Power:**
P = 1/f(m) = 1/1.2 = **+0.833 D**
The combination acts as a **converging lens** of focal length **120 cm** and power **+0.833 Dioptres**.
---
### Q3. Number of Electrons Per Minute
**Given:**
- I = 1.00 mA = 1.00 × 10⁻³ A
- t = 1 minute = 60 s
- e = 1.6 × 10⁻¹⁹ C
**Formula:** Q = It → n = Q/e = It/e
n = (1.00 × 10⁻³ × 60) / (1.6 × 10⁻¹⁹)
n = (6.0 × 10⁻²) / (1.6 × 10⁻¹⁹)
**n = 3.75 × 10¹⁷ electrons**
---
### Q4. Effective Resistance
**Given:**
- Two equal resistors xΩ in parallel, then in series with xΩ
**Step 1 — Two xΩ resistors in parallel:**
R_parallel = (x × x)/(x + x) = x²/2x = **x/2 Ω**
**Step 2 — Result in series with xΩ:**
R_eff = x/2 + x = x/2 + 2x/2
**R_eff = 3x/2 Ω = 1.5x Ω**
---
### Q5. Galvanometer
**Definition:**
A galvanometer is a **sensitive electromagnetic instrument** used to detect and measure very small electric currents. It works on the principle that a current-carrying conductor in a magnetic field experiences a deflecting force.
**(i) Connected in Series:**
When connected in series (with a high resistance — multiplier), it functions as a **voltmeter** — measures **potential difference (voltage)**.
**(ii) Connected in Parallel:**
When connected in parallel (with a low resistance — shunt), it functions as an **ammeter** — measures **electric current**.
---
### Q6. Electric Field Between Parallel Plates
**Given:**
- d = 20 mm = 20 × 10⁻³ m = 0.02 m
- V = 10 V
- e = 1.6 × 10⁻¹⁹ C
**Electric Field Intensity:**
E = V/d = 10 / 0.02
**E = 500 V/m (or 500 N/C)**
**Force on electron:**
F = eE = 1.6 × 10⁻¹⁹ × 500
**F = 8.0 × 10⁻¹⁷ N**
---
### Q7. Force on Electron in Magnetic Field
**Given:**
- v = 1.5 × 10⁷ ms⁻¹
- B = 0.0012 T
- e = 1.6 × 10⁻¹⁹ C
- θ = 90° (perpendicular)
**Formula:** F = Bev sin θ
F = 0.0012 × 1.6 × 10⁻¹⁹ × 1.5 × 10⁷ × sin 90°
F = 0.0012 × 1.6 × 10⁻¹⁹ × 1.5 × 10⁷
F = 0.0012 × 2.4 × 10⁻¹²
**F = 2.88 × 10⁻¹⁵ N**
---
### Q8. Work Function of Metal
**Given:**
- λ = 200 nm = 200 × 10⁻⁹ m
- v_max = 7.22 × 10⁵ ms⁻¹
- h = 6.6 × 10⁻³⁴ Js
- c = 3.0 × 10⁸ ms⁻¹
- mₑ = 9.0 × 10⁻³¹ kg
**Einstein's photoelectric equation:**
hf = φ + ½mv²
**Photon energy:**
E = hc/λ = (6.6 × 10⁻³⁴ × 3.0 × 10⁸) / (200 × 10⁻⁹)
E = 19.8 × 10⁻²⁶ / 2.0 × 10⁻⁷
E = 9.9 × 10⁻¹⁹ J
**Kinetic energy:**
KE = ½mv² = ½ × 9.0 × 10⁻³¹ × (7.22 × 10⁵)²
KE = ½ × 9.0 × 10⁻³¹ × 5.213 × 10¹¹
KE = 2.346 × 10⁻¹⁹ J
**Work function:**
φ = E − KE = 9.9 × 10⁻¹⁹ − 2.346 × 10⁻¹⁹
**φ = 7.554 × 10⁻¹⁹ J**
Converting: φ = 7.554 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = **4.72 eV**
---
### Q9. Nuclear Decay — Values of A and Z
**Given:**
³²₁₅P decays by β⁻ and γ emission → ᴬ_Z X
- **β⁻ decay:** emits an electron (⁰₋₁e) → Z increases by 1, A unchanged
- **γ emission:** emits a gamma photon → no change in A or Z
**Therefore:**
- A = 32 (unchanged)
- Z = 15 + 1 = 16
**A = 32, Z = 16** (the element is ³²₁₆S — Sulfur)
---
### Q10. Decay Constant
**Given:**
- t½ = 23 days = 23 × 24 × 3600 = 1,987,200 s
**Formula:** λ = ln2 / t½ = 0.693 / t½
In days: λ = 0.693 / 23
**λ = 0.0301 day⁻¹**
Or in seconds: λ = 0.693 / 1,987,200 = **3.49 × 10⁻⁷ s⁻¹**
---
## SECTION B: GEOMETRIC OPTICS
### Q11(a) Ray and Beams, Laws of Reflection, Diffuse vs Regular
**i. Ray vs Beam of Monochromatic Light:**
- A **ray** is a single line representing the direction of travel of light — it has no width.
- A **beam** is a collection of rays traveling together:
- **Parallel beam:** rays travel parallel to each other
- **Convergent beam:** rays meet at a point
- **Divergent beam:** rays spread out from a point
*(Labeled ray diagrams should show these in the answer booklet)*
**ii. Laws of Reflection of Red Light:**
1. The **incident ray, reflected ray, and normal** at the point of incidence all lie in the **same plane**.
2. The **angle of incidence (i) equals the angle of reflection (r)**: i = r
*(These laws apply to all colours of light including red)*
**iii. Regular vs Diffuse Reflection:**
- **Regular reflection:** On a **smooth, polished surface** — parallel incident rays produce parallel reflected rays; a clear image is formed. Example: mirror.
- **Diffuse reflection:** On a **rough, uneven surface** — parallel incident rays are reflected in **many random directions**; no clear image formed. Example: wall, paper.
---
### Q11(b) Object at Centre of Curvature of Concave Mirror
**Given:** Object at centre of curvature → u = R = 2f
**Using mirror formula:** 1/f = 1/v + 1/u
Let f = −f (concave), u = −2f (real object)
1/(−f) = 1/v + 1/(−2f)
1/v = 1/(−f) + 1/(2f) = −2/2f + 1/2f = −1/2f
**v = −2f** (image is at centre of curvature, same side as object)
**Magnification:**
m = −v/u = −(−2f)/(−2f) = **−1**
**Comment:** The image is:
- **Real** (v is negative in concave mirror convention)
- **Inverted** (m = −1)
- **Same size as object** (|m| = 1)
- Formed at the **same position as the object** (at centre of curvature)
---
### Q11(c) Object in Front of Concave Mirror
**Given:**
- u = −27 cm (object distance, real)
- f = −18 cm (concave mirror)
**Mirror formula:** 1/f = 1/v + 1/u
1/(−18) = 1/v + 1/(−27)
1/v = −1/18 + 1/27
1/v = −3/54 + 2/54 = −1/54
**v = −54 cm**
The screen should be placed **54 cm** from the mirror.
**Magnification:**
m = −v/u = −(−54)/(−27) = −54/27 = **−2**
**Size of image:** |m| = 2 → image is **twice the size of the object**
**Nature of image:**
- **Real** (formed in front of mirror)
- **Inverted** (m is negative)
- **Magnified** (|m| = 2)
---
### Q12(a) Prism Ray Diagram and Terms
**i. Ray Diagram:** *(Draw in answer booklet)*
A ray enters one face of the prism, refracts, travels through the glass, and exits the other face with further refraction.
**Labels:**
- **Angle of incidence (i):** angle between incident ray and normal at first face
- **Angle of refraction (r):** angle between refracted ray inside prism and normal
- **Angle of emergence (e):** angle between emerging ray and normal at second face
- **Angle of deviation (δ):** angle between original incident ray direction and final emergent ray
- **Angle of prism (A):** apex angle of the prism
**ii. Underlined Terms Explained:**
- **Angle of emergence:** The angle between the ray emerging from the second face of the prism and the normal to that face at the point of emergence.
- **Angle of deviation:** The angle between the direction of the incident ray and the direction of the emergent ray — the total bending of light through the prism.
- **Angle of prism:** The apex angle of the prism — the angle between the two refracting faces.
- **Angle of refraction:** The angle between the refracted ray inside the prism and the normal to the surface at the point of refraction.
**iii. Conditions:**
- **Minimum deviation:** Occurs when the ray passes **symmetrically** through the prism — angle of incidence = angle of emergence (i = e), and the ray inside the prism is **parallel to the base**.
- **Maximum deviation:** Occurs when the angle of incidence is either **very small (grazing incidence)** or **very large**, i.e., at the extreme angles of incidence.
---
### Q12(b-i) Minimum Deviation Formula Reduction
**Given formula:**
μ = sin((A + δₘᵢₙ)/2) / sin(A/2)
**For small angle A**, sin θ ≈ θ (in radians):
μ = (A + δₘᵢₙ)/2 ÷ A/2
μ = (A + δₘᵢₙ) / A
μA = A + δₘᵢₙ
**δₘᵢₙ = μA − A = (μ − 1)A** ✓
**Sketch:** Graph of δ vs i shows a U-shaped curve with minimum at the symmetric ray position.
---
### Q12(b-ii) Angular Dispersion
**Given:**
- A = 6°
- μᵣ (red) = 1.641
- μᵥ (violet) = 1.659
**Angular dispersion:**
Using δ = (μ − 1)A for small angle prism:
δᵣ = (1.641 − 1) × 6° = 0.641 × 6° = 3.846°
δᵥ = (1.659 − 1) × 6° = 0.659 × 6° = 3.954°
**Angular dispersion = δᵥ − δᵣ = 3.954° − 3.846° = 0.108°**
---
### Q12(b-iii) Angle of Incidence for Minimum Deviation
**Given:**
- Equilateral prism: A = 60°
- μ = 1.5
**At minimum deviation:** μ = sin((A + δₘᵢₙ)/2) / sin(A/2)
1.5 = sin((60° + δₘᵢₙ)/2) / sin(30°)
1.5 = sin((60° + δₘᵢₙ)/2) / 0.5
sin((60° + δₘᵢₙ)/2) = 1.5 × 0.5 = 0.75
(60° + δₘᵢₙ)/2 = sin⁻¹(0.75) = 48.59°
60° + δₘᵢₙ = 97.18°
δₘᵢₙ = 37.18°
**At minimum deviation:** i = (A + δₘᵢₙ)/2 = (60° + 37.18°)/2
**i = 48.59° ≈ 48.6°**
---
## SECTION C: ELECTRICITY AND MAGNETISM
### Q13(a)(i) Charging by Friction
When two different **insulating materials** are rubbed together, electrons are transferred from one material to the other due to differences in their **electron affinity**. The material that **gains electrons** becomes **negatively charged**, while the one that **loses electrons** becomes **positively charged**. The total charge is conserved (conservation of charge).
**Example:** Rubbing a **glass rod with silk** — the glass loses electrons to the silk; glass becomes positive (+) and silk becomes negative (−).
---
### Q13(a)(ii) Three Charges A, B, C
**Given:**
- A and B **attract** → they have **opposite charges**
- B and C **repel** → they have the **same charge**
**Determining charges:**
- Since B and C repel → B and C have the **same sign**
- Since A and B attract → A has the **opposite sign** to B
**Possible combinations:**
- If B is **negative** → C is **negative** → A is **positive**
- If B is **positive** → C is **positive** → A is **negative**
**When A and C are brought together:**
- If A = +ve and C = −ve → they **attract**
- If A = −ve and C = +ve → they **attract**
**Conclusion:** A and C will always **attract** each other.
---
### Q13(b) Glass Rod and Silk
**Given:** Charge on glass rod = +19.2 × 10⁻¹⁹ C
**(i) Charge on silk:**
By conservation of charge, the silk acquires an equal and opposite charge:
**Charge on silk = −19.2 × 10⁻¹⁹ C**
**(ii) Number of electrons lost by glass rod:**
n = Q/e = (19.2 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹)
**n = 12 electrons**
**(iii) Force between Na⁺ and Cl⁻ ions:**
**Given:**
- q₁ = q₂ = 1.6 × 10⁻¹⁹ C
- r = 2.8 × 10⁻¹⁰ m
- k = 9 × 10⁹ Nm²C⁻²
F = kq₁q₂/r²
F = (9 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹) / (2.8 × 10⁻¹⁰)²
F = (9 × 10⁹ × 2.56 × 10⁻³⁸) / (7.84 × 10⁻²⁰)
F = (23.04 × 10⁻²⁹) / (7.84 × 10⁻²⁰)
**F = 2.94 × 10⁻⁹ N**
---
### Q14(a) Conductance and Conductivity
**(i) Definitions:**
- **Conductance (G):** The **reciprocal of resistance** — a measure of how easily current flows through a conductor.
G = 1/R; Unit: **Siemens (S)**
- **Conductivity (σ):** The **reciprocal of resistivity (ρ)** — an intrinsic material property indicating how well a material conducts electricity.
σ = 1/ρ; Unit: **Sm⁻¹ or (Ωm)⁻¹**
**(ii) Conductance in terms of conductivity:**
Since R = ρL/A and G = 1/R:
G = 1/R = A/(ρL) = **σA/L**
---
### Q14(b) Resistance of Wire with Different Cross-section
**Given (Wire 1):**
- L = 8 m
- A₁ = 3 mm² = 3 × 10⁻⁶ m²
- R₁ = 0.16 Ω
**Find resistivity:**
ρ = R₁A₁/L = (0.16 × 3 × 10⁻⁶) / 8
ρ = 4.8 × 10⁻⁷ / 8 = **6.0 × 10⁻⁸ Ωm**
**Wire 2 (same length and material):**
- L = 8 m
- A₂ = 1 mm² = 1 × 10⁻⁶ m²
R₂ = ρL/A₂ = (6.0 × 10⁻⁸ × 8) / (1 × 10⁻⁶)
R₂ = 48 × 10⁻⁸ / 10⁻⁶
**R₂ = 0.48 Ω**
---
### Q14(c) Inductance and Energy in Coil
**Given:**
- Φ = 25 mWb = 25 × 10⁻³ Wb
- N = 1500 turns
- I = 3 A
**Inductance:**
L = NΦ/I = (1500 × 25 × 10⁻³) / 3
L = 37.5 / 3
**L = 12.5 H**
**Energy stored:**
E = ½LI² = ½ × 12.5 × 3²
E = ½ × 12.5 × 9
**E = 56.25 J**
---
### Q15(a) Ampere's Law
**Integral form of Ampere's Law:**
∮ B · dl = μ₀I
For a long straight wire, applying Ampere's law over a circular path of radius r:
B × 2πr = μ₀I
**B = μ₀I / 2πr**
---
### Q15(b) Magnetic Field Below Power Line
**Given:**
- r = 7.0 m
- I = 12,000 A
- μ₀ = 4π × 10⁻⁷ NA⁻²
B = μ₀I / 2πr
B = (4π × 10⁻⁷ × 12,000) / (2π × 7.0)
B = (4π × 10⁻⁷ × 12,000) / (14π)
B = (4 × 12,000 × 10⁻⁷) / 14
B = (48,000 × 10⁻⁷) / 14
B = 3428.6 × 10⁻⁷
**B = 3.43 × 10⁻⁴ T**
---
### Q15(c) Capacitor Charge — Before and After
**Given:**
- C₁ = 2μF, V = 12V
- C₂ = 4μF (connected in parallel after disconnection)
**Initial charge on 2μF:**
Q₀ = C₁V = 2 × 10⁻⁶ × 12
**Q₀ = 24 μC**
**After connecting 4μF in parallel:**
Total capacitance: C_total = C₁ + C₂ = 2 + 4 = 6μF
By conservation of charge, total charge remains Q₀ = 24 μC
New voltage: V' = Q₀/C_total = 24/6 = **4 V**
**Final charge on 2μF:**
Q_final = C₁ × V' = 2 × 10⁻⁶ × 4 = **8 μC**
**Summary:**
- Initial charge on 2μF = **24 μC**
- Final charge on 2μF = **8 μC**
---
### Q16(a)(i) Differences Between AC and DC
| Feature | Alternating Current (AC) | Direct Current (DC) |
|---|---|---|
| Direction | Reverses direction periodically | Flows in one direction only |
| Magnitude | Varies sinusoidally with time | Remains constant |
| Frequency | Has a definite frequency (e.g., 50Hz) | Frequency is zero |
| Generation | Generated by alternators | Generated by batteries/cells |
---
### Q16(a)(ii) RC has Units of Time
R has units of **Ω = V/A = V·s/C**
C has units of **F = C/V**
RC = (V·s/C) × (C/V) = **seconds (s)** ✓
---
### Q16(b) Effective Capacitance from Figure 1
**From the circuit:** 50μF and 60μF are in **series** (top branch); 70F, 20μF and 120μF are in **series** (bottom branch); both branches in **parallel**.
**Top branch (50μF and 60μF in series):**
1/C_top = 1/50 + 1/60 = 6/300 + 5/300 = 11/300
C_top = 300/11 = **27.27 μF**
**Bottom branch (70μF, 20μF, 120μF in series):**
*(Note: The 70F appears to be 70μF from context)*
1/C_bot = 1/70 + 1/20 + 1/120
Finding common denominator (840):
= 12/840 + 42/840 + 7/840 = 61/840
C_bot = 840/61 = **13.77 μF**
**Total (parallel combination):**
C_eq = C_top + C_bot = 27.27 + 13.77
**C_eq ≈ 41.04 μF**
---
### Q16(c) Series RLC Circuit — Resonance
**Given:**
- R = 10 Ω
- L = 125 mH = 0.125 H
- C = 60 μF = 60 × 10⁻⁶ F
- V = 120 V
**Resonant frequency:**
f₀ = 1/(2π√LC)
f₀ = 1/(2π√(0.125 × 60 × 10⁻⁶))
f₀ = 1/(2π√(7.5 × 10⁻⁶))
f₀ = 1/(2π × 2.739 × 10⁻³)
f₀ = 1/(1.7214 × 10⁻²)
**f₀ = 58.1 Hz**
**Current at resonance** (Z = R at resonance):
I = V/R = 120/10 = **12 A**
**Q-factor:**
Q = (1/R)√(L/C) = (1/10)√(0.125/60×10⁻⁶)
Q = (1/10)√(2083.3)
Q = (1/10) × 45.64
**Q = 4.564**
---
## SECTION D: MODERN PHYSICS
### Q17(a)(i) Photoelectric Effect and Einstein's Equation
**Concept:**
When light of sufficient frequency strikes a metal surface, electrons are **ejected** from the surface. Key observations:
- Below a **threshold frequency (f₀)**, no electrons are emitted regardless of intensity.
- Above f₀, electrons are emitted **instantaneously**.
- Kinetic energy of emitted electrons depends on **frequency**, not intensity.
- Increasing intensity increases the **number** of electrons, not their energy.
**Einstein's Explanation:**
Light consists of **photons** each with energy E = hf. When a photon hits an electron, it gives all its energy to the electron. Part is used to escape the metal (work function φ), and the rest becomes kinetic energy.
**Einstein's Photoelectric Equation:**
hf = φ + ½mv²_max
Or: hf = hf₀ + ½mv²_max
Where:
- h = Planck's constant
- f = frequency of incident light
- φ = hf₀ = work function
- ½mv²_max = maximum kinetic energy of emitted electron
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### Q17(a)(ii) Equation in Terms of Cutoff Wavelength
Since f = c/λ:
hc/λ = hc/λ₀ + ½mv²
**Graph:** Plot of KE (½mv²) vs frequency f:
- Straight line with slope = **h** (Planck's constant)
- x-intercept = threshold frequency f₀
- y-intercept = −φ (work function)
**From slope:** h = ΔKE/Δf = **slope of graph**
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### Q17(b) Work Function and Planck's Constant
**Given:**
- λ₁ = 4000 Å = 4000 × 10⁻¹⁰ m, V₁ = 2V
- λ₂ = 6000 Å = 6000 × 10⁻¹⁰ m, V₂ = 1V
- e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ ms⁻¹
**Einstein's equation:** hc/λ = φ + eV
**Equation 1:** hc/(4000×10⁻¹⁰) = φ + 2e ...(1)
**Equation 2:** hc/(6000×10⁻¹⁰) = φ + 1e ...(2)
**Subtract (2) from (1):**
hc[1/(4×10⁻⁷) − 1/(6×10⁻⁷)] = e
hc[(3−2)/(12×10⁻⁷)] = e
hc/(12×10⁻⁷) = 1.6×10⁻¹⁹
h = (1.6×10⁻¹⁹ × 12×10⁻⁷) / (3×10⁸)
h = (19.2×10⁻²⁶) / (3×10⁸)
**h = 6.4 × 10⁻³⁴ Js** ✓ (close to standard value)
**Work function from equation (2):**
φ = hc/λ₂ − eV₂
φ = (6.4×10⁻³⁴ × 3×10⁸)/(6×10⁻⁷) − (1.6×10⁻¹⁹ × 1)
φ = (19.2×10⁻²⁶)/(6×10⁻⁷) − 1.6×10⁻¹⁹
φ = 3.2×10⁻¹⁹ − 1.6×10⁻¹⁹
**φ = 1.6 × 10⁻¹⁹ J = 1.0 eV**
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### Q17(c) Energy of Electron Transition (Bohr Model)
**Given:** Eₙ = −13.6/n² eV
**E₃** (n=3): E₃ = −13.6/9 = −1.511 eV
**E₂** (n=2): E₂ = −13.6/4 = −3.4 eV
**Energy released (n=3 → n=2):**
ΔE = E₃ − E₂ = −1.511 − (−3.4)
**ΔE = +1.889 eV ≈ 1.89 eV**
Converting: ΔE = 1.89 × 1.6 × 10⁻¹⁹ = **3.02 × 10⁻¹⁹ J**
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### Q18(a) Half-life and Decay Constant
**Half-life (t½):** The time taken for **half the nuclei** in a radioactive sample to decay. It is a constant for a given nuclide, independent of temperature, pressure, or chemical state.
**Decay constant (λ):** The **probability** that a given nucleus will decay per unit time. It represents the fraction of nuclei decaying per second.
**Deduction of relationship:**
From radioactive decay law: N = N₀e^(−λt)
At t = t½: N = N₀/2
N₀/2 = N₀e^(−λt½)
1/2 = e^(−λt½)
Taking natural log:
ln(1/2) = −λt½
−ln2 = −λt½
**λ = ln2/t½ = 0.693/t½**
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### Q18(b) Decay of Radium-226
**Given:**
- ²²⁶₈₈Ra = 226.0245u
- ²²²₈₆Rn = 222.0175u
- ⁴₂He = 4.0026u
- 1u = 1.66 × 10⁻²⁷ kg
- 1u = 931 MeV
**Decay equation:**
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
**Mass defect (Δm):**
Δm = mass of Ra − (mass of Rn + mass of He)
Δm = 226.0245 − (222.0175 + 4.0026)
Δm = 226.0245 − 226.0201
**Δm = 0.0044 u**
**Energy released:**
E = Δm × 931 MeV/u
E = 0.0044 × 931
**E = 4.096 MeV ≈ 4.10 MeV**
Converting to eV: **≈ 4.10 × 10⁶ eV ≈ 4.99 eV** ✓
This is consistent with the given value of **4.99 eV**, confirming conservation of mass-energy.
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### Q18(c) Uncertainty in Momentum — Heisenberg's Principle
**Given:**
- Δx = 10⁻¹⁴ m (uncertainty in position of proton)
- h = 6.6 × 10⁻³⁴ Js
**Heisenberg Uncertainty Principle:**
Δx · Δp ≥ h/4π
**Minimum uncertainty in momentum:**
Δp = h/(4π·Δx)
Δp = (6.6 × 10⁻³⁴) / (4π × 10⁻¹⁴)
Δp = (6.6 × 10⁻³⁴) / (1.2566 × 10⁻¹³)
**Δp = 5.25 × 10⁻²¹ kg·ms⁻¹**
**Kinetic Energy:**
Using KE = (Δp)²/2m, where m (proton) = 1.67 × 10⁻²⁷ kg:
KE = (5.25 × 10⁻²¹)² / (2 × 1.67 × 10⁻²⁷)
KE = (27.56 × 10⁻⁴²) / (3.34 × 10⁻²⁷)
KE = 8.25 × 10⁻¹⁵ J
Converting: KE = 8.25 × 10⁻¹⁵ / 1.6 × 10⁻¹⁹
**KE ≈ 51,562 eV ≈ 51.6 keV**