FAD1022: Rapid-Fire Drill Pack — Leak Topics

Objective: Master all exam sections based on the detailed leak structure.
Target: 2 min per Section A problem, 5 min per structured problem.
Total problems: ~50
Estimated time: ~120 min

Cheat Sheet (Memorize First)

Physics Constants

Constant	Symbol	Value
Coulomb constant	$k$	$8.99 \times 10^{9}\ \text{N·m}^{2}\text{·C}^{-2}$
Permittivity of free space	$\varepsilon_{0}$	$8.85 \times 10^{-12}\ \text{F·m}^{-1}$
Permeability of free space	$\mu_{0}$	$4\pi \times 10^{-7}\ \text{T·m·A}^{-1}$
Elementary charge	$e$	$1.60 \times 10^{-19}\ \text{C}$
Electron mass	$m_{e}$	$9.11 \times 10^{-31}\ \text{kg}$
Planck's constant	$h$	$6.63 \times 10^{-34}\ \text{J·s}$
Speed of light	$c$	$3.00 \times 10^{8}\ \text{m/s}$
$hc$ (useful)	$hc$	$1240\ \text{eV·nm}$
Bohr radius	$a_{0}$	$0.529\ \text{Å} = 5.29 \times 10^{-11}\ \text{m}$

Electrostatics

Formula	Description
$\vec{F} = k q_1 q_2 / r^{2}\ \hat{r}$	Coulomb's law (vector)
$\vec{E} = k Q / r^{2}\ \hat{r}$	Electric field of point charge
$\vec{F} = q\vec{E}$	Force on charge in field
$a_{y} = qE / m$	Acceleration in uniform E-field
$\vec{E}{\text{net}} = \sum \vec{E}{i}$	Superposition principle
$y = \frac{1}{2} \cdot \frac{qE}{m} \cdot \left(\frac{L}{v_{0}}\right)^{2}$	Vertical deflection in parallel plates
$\theta = \tan^{-1}(v_{y} / v_{x})$	Deflection angle

Capacitors & DC

Formula	Description
$C = Q / \Delta V$	Definition of capacitance
$C = \kappa\varepsilon_{0} A / d$	Parallel-plate with dielectric
$U = \frac{1}{2} C V^{2} = Q^{2} / (2C)$	Stored energy
$\tau = RC$	RC time constant
$q(t) = Q_{0}(1 - e^{-t/\tau})$	Charging capacitor
$q(t) = Q_{0}e^{-t/\tau}$	Discharging capacitor
$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_{2}}{R_{1} + R_{2}}$	Voltage divider (unloaded)
$\sum I_{\text{in}} = \sum I_{\text{out}}$	Kirchhoff's Current Law

AC Circuits & Power

Formula	Description
$V = IZ$	Ohm's law for AC
$X_{L} = \omega L = 2\pi f L$	Inductive reactance
$X_{C} = 1 / (\omega C) = 1 / (2\pi f C)$	Capacitive reactance
$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$	RLC impedance
$\phi = \tan^{-1}\left((X_{L} - X_{C}) / R\right)$	Phase angle
$P_{e} = V_{\text{rms}} I_{\text{rms}} \cos\phi$	Real / average power
$P_{LC} = V_{\text{rms}} I_{\text{rms}} \sin\phi$	Inductive reactive power
$P_{CC} = V_{\text{rms}} I_{\text{rms}} \sin\phi$	Capacitive reactive power
$P_{A} = V_{\text{rms}} I_{\text{rms}}$	Apparent power
$\cos\phi = R / Z$	Power factor
$\omega_{0} = 1 / \sqrt{LC}$	Resonant angular frequency
$f_{0} = 1 / (2\pi\sqrt{LC})$	Resonant frequency

CIVIL Mnemonic: In a Capacitor, I leads V; In an Inductor, V leads I.

Magnetism

Formula	Description
$\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{\text{enc}}$	Ampere's law
$B(r) = \mu_{0} I r / (2\pi R^{2})$	Inside wire ($r < R$)
$B(r) = \mu_{0} I / (2\pi r)$	Outside wire ($r > R$)
$\vec{F}_{B} = q\vec{v} \times \vec{B}$	Lorentz force
$\vert\vec{F}_{B}\vert = \vert q\vert vB\sin\theta$	Lorentz force magnitude
$r = mv / (qB)$	Cyclotron radius
$T = 2\pi m / (qB)$	Period (independent of $v$)
$\tau = NIAB\sin\theta$	Torque on current loop
$\mu = NIA$	Magnetic dipole moment

Transformers

Formula	Description
$V_{s} / V_{p} = N_{s} / N_{p}$	Voltage ratio
$I_{s} / I_{p} = N_{p} / N_{s}$	Current ratio (ideal)
$L = \mu_{0} N^{2} A / l$	Solenoid self-inductance
$V_{L} = -L\ dI/dt$	Inductor EMF
$M = k\sqrt{L_{1}L_{2}}$	Mutual inductance
$U = \frac{1}{2} L I^{2}$	Energy in inductor

Transistors & Biasing

Formula	Description
$I_{E} = I_{B} + I_{C}$	KCL at transistor
$I_{C} = \beta I_{B}$	Current gain
$I_{B} = (V_{CC} - V_{BE}) / R_{B}$	Fixed bias
$I_{B} = (V_{CC} - V_{BE}) / (R_{B} + (\beta+1)R_{E})$	Emitter-stabilized bias
$V_{B} = V_{CC} \cdot R_{2} / (R_{1} + R_{2})$	Voltage divider bias
$V_{CE} = V_{CC} - I_{C}(R_{C} + R_{E})$	Collector-emitter voltage

Atomic & Photoelectric

Formula	Description
$r_{n} = n^{2} a_{0}$	Bohr radius ($n=1,2,3,\dots$)
$E_{n} = -13.6 / n^{2}\ \text{eV}$	Hydrogen energy levels
$E = hf = hc / \lambda$	Photon energy
$K_{\text{max}} = hf - \phi$	Einstein's photoelectric equation
$eV_{s} = K_{\text{max}}$	Stopping potential
$f_{0} = \phi / h$	Threshold frequency
$\lambda_{c} = hc / \phi$	Cutoff wavelength

Amplifier Formulas (Section A)

Configuration	Formula
Inverting op-amp	$V_{\text{out}} = -(R_{f} / R_{\text{in}}), V_{\text{in}}$
Non-inverting op-amp	$V_{\text{out}} = (1 + R_{f} / R_{\text{in}}), V_{\text{in}}$

Section A: Structured Rapid-Fire (15 problems × 2 min)

Target: 2 min per question. Total: ~30 min.

A1–A3: Capacitor Charging & Discharging

A $100\ \mu\text{F}$ capacitor charges through a $10\ \text{k}\Omega$ resistor from a $12\ \text{V}$ battery. What is the time constant $\tau$?
- A) $0.1\ \text{s}$ B) $1.0\ \text{s}$ C) $10\ \text{s}$ D) $100\ \text{s}$
A capacitor discharges through a resistor. After $2\tau$, what fraction of the initial charge remains?
- A) $0.135$ B) $0.368$ C) $0.632$ D) $0.865$
A $47\ \mu\text{F}$ capacitor is connected to a $9\ \text{V}$ battery. How much energy is stored?
- A) $0.95\ \text{mJ}$ B) $1.9\ \text{mJ}$ C) $3.8\ \text{mJ}$ D) $4.2\ \text{mJ}$

A4–A6: Kirchhoff's Current Law

At a junction, $I_{1} = 2\ \text{A}$ flows in, $I_{2} = 5\ \text{A}$ flows out, and $I_{3}$ flows out. What is $I_{3}$?
- A) $7\ \text{A}$ out B) $3\ \text{A}$ out C) $3\ \text{A}$ in D) $-3\ \text{A}$
Three currents meet at a node: $I_{1} = 4\ \text{A}$ (entering), $I_{2} = -3\ \text{A}$ (defined as entering), and $I_{3}$ (defined as leaving). Find $I_{3}$.
- A) $1\ \text{A}$ B) $7\ \text{A}$ C) $-1\ \text{A}$ D) $-7\ \text{A}$
A node has $\sum I = I_{1} + I_{2} + I_{3} = 0$ where $I_{1} = 1.5\ \text{A}$, $I_{2} = -0.5\ \text{A}$. What is $I_{3}$?
- A) $1.0\ \text{A}$ B) $-1.0\ \text{A}$ C) $2.0\ \text{A}$ D) $-2.0\ \text{A}$

A7–A9: Amplifier Formulas

An inverting op-amp has $R_{f} = 100\ \text{k}\Omega$ and $R_{\text{in}} = 10\ \text{k}\Omega$. If $V_{\text{in}} = 0.5\ \text{V}$, what is $V_{\text{out}}$?
- A) $+5.5\ \text{V}$ B) $-5.0\ \text{V}$ C) $+5.0\ \text{V}$ D) $-0.5\ \text{V}$
A non-inverting op-amp has $R_{f} = 47\ \text{k}\Omega$ and $R_{\text{in}} = 4.7\ \text{k}\Omega$. What is the voltage gain?
- A) $10$ B) $11$ C) $-10$ D) $-11$
An inverting amplifier has $V_{\text{out}} = -3.0\ \text{V}$ and $V_{\text{in}} = 0.25\ \text{V}$. If $R_{\text{in}} = 10\ \text{k}\Omega$, find $R_{f}$.
- A) $30\ \text{k}\Omega$ B) $60\ \text{k}\Omega$ C) $120\ \text{k}\Omega$ D) $7.5\ \text{k}\Omega$

A10–A12: Ampere's Law — Long Wire $r<R$

A cylindrical wire of radius $R = 2\ \text{mm}$ carries $I = 10\ \text{A}$. What is $B$ at $r = 1\ \text{mm}$ from the centre?
- A) $5.0 \times 10^{-4}\ \text{T}$ B) $2.5 \times 10^{-4}\ \text{T}$ C) $1.0 \times 10^{-3}\ \text{T}$ D) $1.25 \times 10^{-3}\ \text{T}$
For the same wire, what is $B$ at the surface ($r = R$)?
- A) $5.0 \times 10^{-4}\ \text{T}$ B) $1.0 \times 10^{-3}\ \text{T}$ C) $2.0 \times 10^{-3}\ \text{T}$ D) $4.0 \times 10^{-4}\ \text{T}$
Inside a current-carrying wire, $B(r)$ is proportional to:
- A) $r$ B) $1/r$ C) $r^{2}$ D) constant

A13–A15: Mixed Topics

Amplifier: A summing amplifier has two inputs: $V_{1} = 1\ \text{V}$ and $V_{2} = 2\ \text{V}$, with $R_{1} = R_{2} = R_{f} = 10\ \text{k}\Omega$. What is $V_{\text{out}}$?
- A) $-3\ \text{V}$ B) $+3\ \text{V}$ C) $-1.5\ \text{V}$ D) $+1.5\ \text{V}$
Capacitor: A $10\ \mu\text{F}$ capacitor charged to $20\ \text{V}$ is disconnected and then connected across an uncharged $10\ \mu\text{F}$ capacitor. The final voltage is:
- A) $20\ \text{V}$ B) $10\ \text{V}$ C) $5\ \text{V}$ D) $40\ \text{V}$
Kirchhoff + RC: In an RC charging circuit at $t = \tau$, the capacitor voltage is what fraction of the source voltage?
- A) $0.368$ B) $0.500$ C) $0.632$ D) $0.865$

Score: ___/15

Section B: Structured Questions (12 structured problems × ~5 min each)

Exam format: Pick 3 of 4 questions. Each question has 3 parts totalling 12 marks.
Target: ~18 min per full question (6 min per part).

B1: Electrostatics — E-field Vector & Projectile Motion

Part (a) — Electric Field Superposition (4 marks)

Two point charges are placed on the $x$-axis: $Q_{1} = +4\ \mu\text{C}$ at $x = 0$, and $Q_{2} = -9\ \mu\text{C}$ at $x = 5\ \text{cm}$. Calculate the net electric field $\vec{E}_{\text{net}}$ at $x = 2\ \text{cm}$ (magnitude and direction).

Part (b) — Force on Test Charge (3 marks)

A test charge $q = +2\ \text{nC}$ is placed at $x = 2\ \text{cm}$. Calculate the electrostatic force $\vec{F}$ acting on it.

Part (c) — Projectile in Uniform E-field (5 marks)

An electron ($q = -e$, $m = 9.11 \times 10^{-31}\ \text{kg}$) enters horizontally at $v_{0} = 3.0 \times 10^{6}\ \text{m/s}$ into a uniform electric field $E = 5000\ \text{N/C}$ directed downward between parallel plates of length $L = 6.0\ \text{cm}$.

(i) Find the vertical acceleration $a_{y}$ of the electron.
(ii) How long does the electron take to traverse the plates?
(iii) Determine the vertical deflection $y$ of the electron as it exits the plates.
(iv) Find the exit angle $\theta$ relative to the horizontal.

B2: Capacitor + Dielectric + DC Circuit

Part (a) — Capacitance with Dielectric (3 marks)

A parallel-plate capacitor has plates of area $A = 50\ \text{cm}^{2}$ separated by $d = 0.20\ \text{mm}$ in air. A dielectric slab with $\kappa = 3.5$ is inserted filling the gap. Calculate the capacitance $C$.

Part (b) — Energy Stored (4 marks)

The capacitor from part (a) is connected to a $24\ \text{V}$ battery.

(i) Calculate the charge $Q$ on each plate.
(ii) Calculate the energy $U$ stored in the capacitor.
(iii) While the battery remains connected, the dielectric is removed. Explain what happens to the charge and energy. Calculate the new stored energy.

Part (c) — Voltage Divider (5 marks)

A voltage divider consists of $R_{1} = 10\ \text{k}\Omega$ and $R_{2} = 15\ \text{k}\Omega$ connected across a $12\ \text{V}$ supply.

(i) Calculate the unloaded output voltage $V_{\text{out}}$ across $R_{2}$.
(ii) A load $R_{L} = 5\ \text{k}\Omega$ is connected across $R_{2}$. Find the new output voltage.
(iii) By what percentage did the output voltage drop due to the loading effect?

B3: AC Circuits — Phasor Diagrams & Power Types

Part (a) — Phasor Diagram (4 marks)

An RLC series circuit has $R = 40\ \Omega$, $X_{L} = 70\ \Omega$, $X_{C} = 30\ \Omega$, and $I_{\text{rms}} = 2.0\ \text{A}$.

(i) Calculate $V_{R}$, $V_{L}$, and $V_{C}$.
(ii) Draw the phasor diagram showing all three voltages and the resultant $V_{T}$.
(iii) Calculate the magnitude of $V_{T}$ and the phase angle $\phi$.

Part (b) — Power Calculations (4 marks)

For the same circuit:

(i) Calculate real power $P_{e}$, inductive reactive power $P_{LC}$, and capacitive reactive power $P_{CC}$.
(ii) Calculate the apparent power $P_{A}$ and power factor $\cos\phi$.
(iii) Is the circuit inductive or capacitive? Explain.

Part (c) — Pure Circuit Power (4 marks)

A $50\ \mu\text{F}$ capacitor is connected across a $240\ \text{V}_{\text{rms}}$, $50\ \text{Hz}$ supply.

(i) Calculate $X_{C}$ and the rms current.
(ii) Find $P_{e}$, $P_{CC}$, and $P_{A}$.
(iii) What is the phase angle $\phi$ for a pure capacitive circuit? Use CIVIL to justify.

B4: AC Claims & RLC Resonance

Part (a) — Spot the Wrong Claims (4 marks)

Four statements about AC circuits are given. Identify two that are wrong and rewrite them correctly.

"At resonance in a series RLC circuit, the impedance is maximum."
"In a pure inductor, the voltage leads the current by $90^{\circ}$."
"The power factor at resonance is zero."
"Adding a capacitor to an inductive circuit always decreases the power factor."

Part (b) — Identify Parallel Components (3 marks)

A circuit diagram shows a $60\ \Omega$ resistor in series with a parallel combination of a $0.50\ \text{H}$ inductor and a $20\ \mu\text{F}$ capacitor, connected to a $120\ \text{V}_{\text{rms}}$, $60\ \text{Hz}$ supply.

(i) Identify which two components are in parallel.
(ii) Calculate $X_{L}$ and $X_{C}$ at $60\ \text{Hz}$.
(iii) Is this circuit at resonance? Justify.

Part (c) — Make the Circuit Resonate (5 marks)

An RLC series circuit has $R = 20\ \Omega$, $L = 4.0\ \text{mH}$, and $C = 0.50\ \mu\text{F}$. The source frequency is $f = 5.0\ \text{kHz}$.

(i) Calculate $X_{L}$ and $X_{C}$ at $5.0\ \text{kHz}$. Is the circuit inductive or capacitive?
(ii) Find the resonant frequency $f_{0}$ of this circuit.
(iii) To achieve resonance at $5.0\ \text{kHz}$, which component value must change? Calculate the new value.

Score: ___/12 per question (B1–B4).
Total Section B (best 3): ___/36

Section C: Structured Questions (16 structured problems × ~5 min each)

Exam format: Pick 4 of 6 questions. Each question has 2–4 parts totalling 12 marks.
Target: ~18 min per full question.

C1: Magnetism — Ampere's Law & Gauss

Part (a) — Ampere's Law Derivation (5 marks)

A long straight cylindrical wire of radius $R$ carries a uniform current density $J = I/(\pi R^{2})$.

(i) State Ampere's law in integral form.
(ii) Derive the magnetic field $B(r)$ at a distance $r < R$ from the centre. Start by choosing the appropriate Amperian loop. Show all steps clearly.
(iii) Derive $B(r)$ for $r > R$.

Part (b) — 4-Wire System (4 marks)

Four long parallel wires are arranged at the corners of a square of side $a = 10\ \text{cm}$. Each wire carries current $I = 5\ \text{A}$. Wires at the bottom-left and top-right corners carry current into the page ($\otimes$). Wires at the bottom-right and top-left corners carry current out of the page ($\odot$).

(i) Sketch the arrangement and label current directions.
(ii) Calculate the net magnetic field $\vec{B}_{\text{net}}$ at the centre of the square (magnitude and direction).

Part (c) — Gauss's Law for Magnetism (3 marks)

(i) State Gauss's law for magnetism in integral form.
(ii) Explain its physical meaning — what does it tell us about magnetic monopoles?
(iii) A magnetic dipole is placed at the centre of a sphere of radius $r$. What is the net magnetic flux through the spherical surface? Justify your answer.

C2: EM — AC Motor Torque

Part (a) — Torque Derivation & Calculation (6 marks)

A rectangular coil of $N = 100$ turns, width $w = 4.0\ \text{cm}$, and height $h = 6.0\ \text{cm}$ carries current $I = 2.5\ \text{A}$. It is placed in a uniform magnetic field $B = 0.40\ \text{T}$.

(i) Derive the expression $\tau = NIAB\sin\theta$. Define each symbol.
(ii) Calculate the magnetic dipole moment $\mu$ of the coil.
(iii) Calculate the maximum torque on the coil and state the orientation that produces it.

Part (b) — AC Motor Operation (6 marks)

The coil from part (a) is used in an AC motor with angular frequency $\omega = 120\pi\ \text{rad/s}$ ($60\ \text{Hz}$).

(i) Write the expression for instantaneous torque $\tau(t)$.
(ii) Calculate the average torque over one full cycle. (Hint: $\tau_{\text{avg}} = \frac{2}{\pi} NIAB$)
(iii) Explain why the torque direction remains consistent throughout the cycle in an AC motor.

C3: Transformer — Self & Mutual Induction

Part (a) — Self-Inductance of a Solenoid (4 marks)

A solenoid of length $l = 0.30\ \text{m}$, cross-sectional area $A = 4.0\ \text{cm}^{2}$, and $N = 500$ turns carries a current $I = 3.0\ \text{A}$.

(i) Derive the self-inductance $L$ of the solenoid. Start from $B = \mu_{0} n I$.
(ii) Calculate the numerical value of $L$ in $\text{mH}$.
(iii) Calculate the energy stored in the solenoid's magnetic field.

Part (b) — Mutual Inductance (4 marks)

Two coils are placed near each other. Coil 1 has $L_{1} = 5.0\ \text{mH}$, coil 2 has $L_{2} = 20\ \text{mH}$. The coupling coefficient is $k = 0.80$.

(i) Calculate the mutual inductance $M$ between the coils.
(ii) If the current in coil 1 changes at $dI_{1}/dt = 200\ \text{A/s}$, what is the induced EMF in coil 2?
(iii) State the reciprocity principle for mutual inductance.

Part (c) — Ideal Transformer (4 marks)

An ideal transformer has a primary coil with $N_{p} = 200$ turns and a secondary coil with $N_{s} = 800$ turns. The primary is connected to $240\ \text{V}{\text{rms}}$, $50\ \text{Hz}$ and draws $I{p} = 0.50\ \text{A}$.

(i) Determine whether this is a step-up or step-down transformer.
(ii) Calculate $V_{s}$ and $I_{s}$.
(iii) A resistive load is connected to the secondary. Calculate the load resistance $R_{L}$.

C4: Semiconductor & Biasing

Part (a) — Clipper / Rectifier Waveform (3 marks)

A half-wave rectifier circuit uses a silicon diode ($V_{\gamma} \approx 0.7\ \text{V}$) with an input $V_{\text{in}}(t) = 10\sin(\omega t)$.

(i) Sketch the input and output waveforms for one full cycle. Label all voltages.
(ii) At what input voltage does the diode begin conducting?
(iii) What is the peak output voltage?

Part (b) — Voltage Divider Bias Stability (3 marks)

A voltage divider bias circuit uses $R_{1} = 56\ \text{k}\Omega$, $R_{2} = 12\ \text{k}\Omega$, $R_{C} = 2.2\ \text{k}\Omega$, $R_{E} = 1.0\ \text{k}\Omega$, and $V_{CC} = 15\ \text{V}$.

(i) Calculate $V_{B}$ (base voltage).
(ii) Calculate $V_{E}$, $I_{E}$, and $I_{C}$ (assume $V_{BE} = 0.7\ \text{V}$ and $I_{C} \approx I_{E}$).
(iii) Explain why voltage divider bias is more stable than fixed bias with respect to $\beta$ variations.

Part (c) — $I_{E} = I_{B} + I_{C}$ Relation (3 marks)

(i) Prove, using KCL, that $I_{E} = I_{B} + I_{C}$ at the transistor.
(ii) If $\beta = 150$ and $I_{B} = 20\ \mu\text{A}$, find $I_{C}$ and $I_{E}$.
(iii) Show that $I_{E} = (\beta + 1)I_{B}$.

Part (d) — Beta Independence Calculation (3 marks)

An emitter-stabilized bias circuit has $V_{CC} = 20\ \text{V}$, $R_{B} = 470\ \text{k}\Omega$, $R_{C} = 3.3\ \text{k}\Omega$, $R_{E} = 1.5\ \text{k}\Omega$, $V_{BE} = 0.7\ \text{V}$.

(i) Calculate $I_{B}$, $I_{C}$, and $V_{CE}$ for $\beta = 100$.
(ii) Recalculate $I_{C}$ for $\beta = 200$ (all other values unchanged).
(iii) By what percentage did $I_{C}$ change? Discuss the stability.

C5: Atomic Physics — Radius Derivation

Part (a) — Bohr Radius Derivation (6 marks)

(i) State Bohr's three postulates for the hydrogen atom.
(ii) Derive the expression for the radius of the $n$th Bohr orbit: $$r_{n} = \frac{4\pi\varepsilon_{0} n^{2} \hbar^{2}}{m_{e} e^{2}}$$ Start from Coulomb force = centripetal force and angular momentum quantization. Show all steps.
(iii) Calculate the numerical value of the Bohr radius $a_{0}$ (for $n=1$) using the constants provided.

Part (b) — Energy Levels and Transitions (6 marks)

(i) Calculate the energy $E_{n}$ for $n=1$, $2$, $3$, and $\infty$ in eV for hydrogen.
(ii) A hydrogen atom transitions from $n=3$ to $n=2$. Find the energy and wavelength of the emitted photon. Use $hc = 1240\ \text{eV·nm}$.
(iii) What series does this transition belong to? Is this visible light? (Visible range: $\approx 400 - 700\ \text{nm}$)

C6: Photoelectric Effect (Past 3 Years Style)

Part (a) — Photon Energy & Threshold (4 marks)

A metal surface has a work function $\phi = 2.30\ \text{eV}$. Light of wavelength $\lambda = 450\ \text{nm}$ is incident on it.

(i) Calculate the photon energy in eV and in Joules. Use $hc = 1240\ \text{eV·nm}$.
(ii) Calculate the threshold frequency $f_{0}$ and cutoff wavelength $\lambda_{c}$.
(iii) Will photoelectrons be emitted? Justify.

Part (b) — Kinetic Energy & Stopping Potential (4 marks)

(i) Calculate the maximum kinetic energy $K_{\text{max}}$ of the emitted electrons in eV and Joules.
(ii) Find the stopping potential $V_{s}$.
(iii) Calculate the maximum speed $v_{\text{max}}$ of the ejected electrons. ($m_{e} = 9.11 \times 10^{-31}\ \text{kg}$)

Part (c) — Graph Interpretation & Variation (4 marks)

(i) Sketch a graph of $K_{\text{max}}$ vs. frequency $f$. Label the slope, $y$-intercept, and $x$-intercept. What physical quantities do these represent?
(ii) If the light intensity is doubled while keeping the same frequency, what happens to: (a) $K_{\text{max}}$, (b) the photocurrent?
(iii) If a different metal with a larger work function is used, how does the $K_{\text{max}}$ vs. $f$ graph change?

Score: ___/12 per question (C1–C6).
Total Section C (best 4): ___/48

Part D: Mixed & Reverse Engineering (5 problems × 3 min)

Target: 3 min per problem. These combine multiple topics from the exam.

D1 — Mixed Problems

Reverse — Capacitor: An RC circuit has $\tau = 2.5\ \text{s}$. If $R = 50\ \text{k}\Omega$, find $C$. For this $RC$ pair, how long does it take to charge to $90%$ of maximum?
Reverse — Ampere: At $r = 0.5\ \text{mm}$ inside a cylindrical wire, $B = 3.0 \times 10^{-4}\ \text{T}$. The wire radius is $R = 2.0\ \text{mm}$. Find the total current $I$ in the wire.
Multi-concept — Transformer & AC: A step-down transformer ($N_{p}:N_{s} = 10:1$) is connected to $240\ \text{V}{\text{rms}}$, $50\ \text{Hz}$. The secondary feeds a $12\ \Omega$ resistive load. Find $V{s}$, $I_{s}$, $I_{p}$, and the power delivered to the load.
Multi-concept — Transistor & Voltage Divider: A voltage divider bias circuit has $V_{CC} = 12\ \text{V}$, $R_{1} = 33\ \text{k}\Omega$, $R_{2} = 10\ \text{k}\Omega$, $R_{C} = 2.0\ \text{k}\Omega$, $R_{E} = 1.0\ \text{k}\Omega$, $\beta = 120$, $V_{BE} = 0.7\ \text{V}$. Find the Q-point ($I_{C}$, $V_{CE}$). If $\beta$ doubles to $240$, find the new $I_{C}$ and comment on stability.
Multi-concept — Photoelectric + Atomic: A hydrogen atom in its ground state ($n=1$, $E_{1} = -13.6\ \text{eV}$) is struck by a photon of wavelength $\lambda = 91.2\ \text{nm}$. (i) Calculate the photon energy in eV. (ii) Will the atom be ionized? (iii) If the electron is ejected, find its $K_{\text{max}}$. (Hint: $E_{\infty} = 0$)

Score: ___/5

Final Scorecard

Section	Problems	Raw Score
A — Structured (all compulsory)	15	___/15
B — Structured (best 3 of 4)	12 (across 3 Qs)	___/36
C — Structured (best 4 of 6)	16 (across 4 Qs)	___/48
D — Mixed & Reverse	5	___/5
TOTAL	~48	___/104

Proficiency Benchmarks

Level	Score	Meaning
Exam-Ready	$\ge 88/104$ (85%)	Any mistake is a careless slip. Focus on speed.
Solid	$73-87/104$ (70-84%)	Good on most patterns. Target specific weak areas.
Developing	$52-72/104$ (50-69%)	Re-drill the sections you scored lowest on.
Needs Work	$< 52/104$ (<50%)	Study the Cheat Sheet again, then re-attempt.

Speed Benchmarks

<90 min: Excellent mechanical fluency.
90–120 min: Good. Review missed patterns.
>120 min: Re-drill specific weak sections tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem	Reason
e.g. 1	time constant formula

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Section A — Structured

B — $\tau = RC = 10 \times 10^{3} \times 100 \times 10^{-6} = 1.0\ \text{s}$
A — $Q/Q_{0} = e^{-2} \approx 0.135$
B — $U = \frac{1}{2} CV^{2} = \frac{1}{2}(47 \times 10^{-6})(9)^{2} = 1.90 \times 10^{-3}\ \text{J} = 1.9\ \text{mJ}$
C — KCL: $2 = 5 + I_{3} \Rightarrow I_{3} = -3\ \text{A}$ (i.e., $3\ \text{A}$ flowing inward)
A — KCL with sign convention: $4 + (-3) = I_{3} \Rightarrow I_{3} = 1\ \text{A}$
B — $1.5 + 0.5 + I_{3} = 0$ (wait, that gives $2.0 + I_{3} = 0 \Rightarrow I_{3} = -2.0$). Actually: $1.5 + (-0.5) + I_{3} = 0 \Rightarrow I_{3} = -1.0\ \text{A}$. So B.
B — $V_{\text{out}} = -(100/10)(0.5) = -5.0\ \text{V}$
B — Gain $= 1 + 47/4.7 = 1 + 10 = 11$
C — $-3.0 = -(R_{f}/10)(0.25) \Rightarrow R_{f} = 3.0 \times 10 / 0.25 = 120\ \text{k}\Omega$
A — $B = \mu_{0} I r / (2\pi R^{2}) = (4\pi \times 10^{-7})(10)(0.001) / (2\pi \times 0.002^{2}) = 4\pi \times 10^{-6} / (2\pi \times 4 \times 10^{-6}) \times 10 = \dots$ Let me compute: $B = (4\pi \times 10^{-7} \times 10 \times 0.001) / (2\pi \times (0.002)^{2}) = (4\pi \times 10^{-9}) / (2\pi \times 4 \times 10^{-6}) = (4\pi \times 10^{-9}) / (8\pi \times 10^{-6}) = 0.5 \times 10^{-3} = 5.0 \times 10^{-4}\ \text{T}$.
B — At $r = R$: $B = \mu_{0} I / (2\pi R) = (4\pi \times 10^{-7})(10) / (2\pi \times 0.002) = (4\pi \times 10^{-6}) / (2\pi \times 0.002) = (4\pi \times 10^{-6}) / (4\pi \times 10^{-3}) = 1.0 \times 10^{-3}\ \text{T}$. Wait: $(4\pi \times 10^{-7})(10) = 4\pi \times 10^{-6}$. Denominator: $2\pi \times 0.002 = 4\pi \times 10^{-3}$. So $B = (4\pi \times 10^{-6}) / (4\pi \times 10^{-3}) = 10^{-3}\ \text{T}$.
A — $B(r) \propto r$ inside the wire
A — For inverting summing amp: $V_{\text{out}} = -R_{f}(V_{1}/R_{1} + V_{2}/R_{2}) = -10(1/10 + 2/10) = -3\ \text{V}$
B — Charge redistribution: $Q_{\text{total}} = CV = (10 \times 10^{-6})(20) = 200\ \mu\text{C}$. After connecting: $C_{\text{eq}} = 20\ \mu\text{F}$, $V = 200/20 = 10\ \text{V}$.
C — $V_{C}/V_{0} = 1 - e^{-1} \approx 0.632$

Section B — B1 Electrostatics

Part (a): $E_{1}$ from $Q_{1}$ at $x=2$: $r_{1} = 0.02\ \text{m}$, $E_{1} = k|Q_{1}|/r_{1}^{2} = (9\times10^{9})(4\times10^{-6})/(0.02)^{2} = (9\times10^{9})(4\times10^{-6})/(4\times10^{-4}) = 9\times10^{7}\ \text{N/C}$, direction: away from $+$, so $+x$.

$E_{2}$ from $Q_{2}$ at $x=2$: $r_{2}=0.03\ \text{m}$, $E_{2}=k|Q_{2}|/r_{2}^{2} = (9\times10^{9})(9\times10^{-6})/(9\times10^{-4}) = 9\times10^{7}\ \text{N/C}$, direction: toward $-$ charge at $x=5$, which is also $+x$ (since the field at $x=2$ due to a negative charge at $x=5$ is toward the charge, i.e., rightward).

$E_{\text{net}} = E_{1} + E_{2} = 2 \times 9\times10^{7} = 1.8 \times 10^{8}\ \text{N/C}$, direction $+x$.

Part (b): $F = qE = (2\times10^{-9})(1.8\times10^{8}) = 0.36\ \text{N}$, direction $+x$.

Part (c): (i) $F_{y} = qE = (-1.6\times10^{-19})(5000) = -8.0\times10^{-16}\ \text{N}$. $a_{y}=F_{y}/m = -8.0\times10^{-16} / 9.11\times10^{-31} = -8.78\times10^{14}\ \text{m/s}^{2}$ (upward, since electron is negative and field is downward).

(ii) $t = L/v_{0} = 0.06 / (3.0\times10^{6}) = 2.0\times10^{-8}\ \text{s} = 20\ \text{ns}$.

(iii) $y = \frac{1}{2} a_{y} t^{2} = \frac{1}{2}(-8.78\times10^{14})(2.0\times10^{-8})^{2} = \frac{1}{2}(-8.78\times10^{14})(4.0\times10^{-16}) = \frac{1}{2}(-0.351) = -0.176\ \text{m}$ (upward deflection of $17.6\ \text{cm}$ — but check: plates probably aren't this long, may need to check if this exceeds plate separation; the question didn't give plate separation so it's assumed to fit).

(iv) $v_{y} = a_{y}t = (-8.78\times10^{14})(2.0\times10^{-8}) = -1.76\times10^{7}\ \text{m/s}$. $\theta = \tan^{-1}(v_{y}/v_{x}) = \tan^{-1}(-1.76\times10^{7} / 3.0\times10^{6}) = \tan^{-1}(-5.87) \approx -80.3^{\circ}$ (below the horizontal).

Section B — B2 Capacitor + Dielectric + DC

Part (a): $C = \kappa\varepsilon_{0}A/d = (3.5)(8.85\times10^{-12})(50\times10^{-4}) / (0.20\times10^{-3}) = (3.5)(8.85\times10^{-12})(5.0\times10^{-3}) / (2.0\times10^{-4}) = 3.5 \times 8.85 \times 10^{-12} \times 25 = 7.74\times10^{-10}\ \text{F} \approx 774\ \text{pF}$.

Part (b): (i) $Q = CV = (7.74\times10^{-10})(24) = 1.86\times10^{-8}\ \text{C} \approx 18.6\ \text{nC}$. (ii) $U = \frac{1}{2} CV^{2} = \frac{1}{2}(7.74\times10^{-10})(24)^{2} = \frac{1}{2}(7.74\times10^{-10})(576) = 2.23\times10^{-7}\ \text{J} \approx 223\ \text{nJ}$. (iii) Battery connected so $V$ constant. Removing dielectric: $C$ decreases to $C_{0} = C/\kappa = 7.74\times10^{-10}/3.5 = 2.21\times10^{-10}\ \text{F}$. $Q$ decreases to $Q_{0} = C_{0}V = 5.30\times10^{-9}\ \text{C}$. $U_{0} = \frac{1}{2}C_{0}V^{2} = \frac{1}{2}(2.21\times10^{-10})(576) = 6.37\times10^{-8}\ \text{J}$. Energy decreased because the battery took back charge.

Part (c): (i) $V_{\text{out}} = 12 \times 15/(10+15) = 12 \times 15/25 = 7.2\ \text{V}$. (ii) $R_{\text{eq}} = R_{2} \parallel R_{L} = (15\times5)/(15+5) = 75/20 = 3.75\ \text{k}\Omega$. $V_{\text{out}} = 12 \times 3.75/(10+3.75) = 12 \times 3.75/13.75 \approx 3.27\ \text{V}$. (iii) Percentage drop $= (7.2 - 3.27)/7.2 \times 100% \approx 54.6%$.

Section B — B3 AC Phasor & Power

Part (a): (i) $V_{R} = IR = 2.0 \times 40 = 80\ \text{V}$. $V_{L} = IX_{L} = 2.0 \times 70 = 140\ \text{V}$. $V_{C} = IX_{C} = 2.0 \times 30 = 60\ \text{V}$. (ii) Phasor diagram: $V_{R}$ along $0^{\circ}$ (right), $V_{L}$ at $+90^{\circ}$ (up), $V_{C}$ at $-90^{\circ}$ (down). Net vertical: $140 - 60 = 80\ \text{V}$ up. (iii) $V_{T} = \sqrt{80^{2} + 80^{2}} = 80\sqrt{2} \approx 113\ \text{V}$. $\phi = \tan^{-1}(80/80) = 45^{\circ}$ (voltage leads current).

Part (b): (i) $P_{e} = I^{2}R = (2.0)^{2}(40) = 160\ \text{W}$. $P_{LC} = I^{2}X_{L} = 4.0 \times 70 = 280\ \text{VAR}$. $P_{CC} = I^{2}X_{C} = 4.0 \times 30 = 120\ \text{VAR}$. (ii) $P_{A} = IV_{T} = 2.0 \times 113 = 226\ \text{VA}$. $\cos\phi = P_{e}/P_{A} = 160/226 \approx 0.708$. (iii) Inductive, because $X_{L} > X_{C}$, so $\phi > 0$, voltage leads current.

Part (c): (i) $X_{C} = 1/(2\pi f C) = 1/(2\pi \times 50 \times 50\times10^{-6}) = 1/(0.0157) \approx 63.7\ \Omega$. $I_{\text{rms}} = V/X_{C} = 240/63.7 \approx 3.77\ \text{A}$. (ii) $P_{e} = 0$ (pure capacitive). $P_{CC} = V_{\text{rms}} I_{\text{rms}} = 240 \times 3.77 \approx 905\ \text{VAR}$. $P_{A} = 905\ \text{VA}$. (iii) $\phi = -90^{\circ}$. By CIVIL: In a Capacitor, I leads V by $90^{\circ}$.

Section B — B4 AC Claims & RLC

Part (a): Wrong statements: 1 and 3 and 4 (pick any two).

Wrong. Correct: "At resonance in a series RLC circuit, the impedance is minimum ($Z=R$)."
Wrong. Correct: "The power factor at resonance is unity ($\cos\phi=1$)."
Wrong. Correct: "Adding a capacitor to an inductive circuit increases the power factor (PF correction)."

Part (b): (i) The inductor ($0.50\ \text{H}$) and capacitor ($20\ \mu\text{F}$) are in parallel with each other. (ii) $X_{L} = 2\pi(60)(0.50) = 188.5\ \Omega$. $X_{C} = 1/(2\pi(60)(20\times10^{-6})) = 1/(0.00754) \approx 132.6\ \Omega$. $X_{L} \neq X_{C}$, so not at resonance. (iii) Not at resonance because $X_{L} \neq X_{C}$.

Part (c): (i) $X_{L} = 2\pi(5000)(0.004) = 125.7\ \Omega$. $X_{C} = 1/(2\pi(5000)(0.5\times10^{-6})) = 1/(0.0157) \approx 63.7\ \Omega$. Inductive ($X_{L} > X_{C}$). (ii) $f_{0} = 1/(2\pi\sqrt{LC}) = 1/(2\pi\sqrt{4\times10^{-3} \times 0.5\times10^{-6}}) = 1/(2\pi\sqrt{2\times10^{-9}}) = 1/(2\pi \times 4.47\times10^{-5}) \approx 3560\ \text{Hz}$. (iii) At $f=5.0\ \text{kHz}$, need $X_{L} = X_{C}$. Either change $L$ to $L = 1/(\omega^{2}C) = 1/((2\pi\times5000)^{2} \times 0.5\times10^{-6}) = 1/(9.87\times10^{8} \times 0.5\times10^{-6}) = 1/(493.5) \approx 2.03\ \text{mH}$. Or change $C$ to $C = 1/(\omega^{2}L) = 1/((2\pi\times5000)^{2} \times 4\times10^{-3}) = 1/(3.95\times10^{6}) \approx 0.253\ \mu\text{F}$.

Section C — C1 Magnetism

Part (a): (i) $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{\text{enc}}$. (ii) Amperian loop: circle of radius $r$ ($r < R$). By symmetry $B$ is tangential and constant. $\oint B dl = B \cdot 2\pi r$. Current enclosed: $I_{\text{enc}} = J \cdot \pi r^{2} = (I/\pi R^{2})\pi r^{2} = Ir^{2}/R^{2}$. So $B \cdot 2\pi r = \mu_{0} I r^{2}/R^{2} \Rightarrow B = \mu_{0} I r / (2\pi R^{2})$. (iii) For $r > R$: $I_{\text{enc}} = I$. So $B \cdot 2\pi r = \mu_{0} I \Rightarrow B = \mu_{0} I / (2\pi r)$.

Part (b): (i) & (ii) At the centre of the square, each wire contributes $B = \mu_{0}I/(2\pi r)$ where $r = a/\sqrt{2} = 0.10/\sqrt{2} = 0.0707\ \text{m}$. $B_{\text{each}} = (4\pi\times10^{-7})(5)/(2\pi \times 0.0707) = (2\times10^{-6})(5)/0.0707 = 1.414\times10^{-4}\ \text{T}$. By symmetry: fields from opposite-corner wires add constructively along the diagonal direction. The two diagonals are perpendicular. Net field: $B_{\text{net}} = 2\sqrt{2} \times B_{\text{each}} \approx 2.828 \times 1.414\times10^{-4} = 4.0\times10^{-4}\ \text{T}$. Direction: at $45^{\circ}$ to the horizontal.

Part (c): (i) $\oint \vec{B} \cdot d\vec{A} = 0$. (ii) It means the net magnetic flux through any closed surface is zero. This implies magnetic monopoles do not exist — magnetic field lines form closed loops. (iii) Zero. By Gauss's law for magnetism, the net flux through any closed surface is always zero, regardless of what is inside.

Section C — C2 AC Motor Torque

Part (a): (i) Force on each side of loop: $F = ILB\sin\theta$ where $I$ = current, $L$ = length of side in field, $B$ = field, $\theta$ = angle. For a rectangular loop, two sides experience equal and opposite forces. Torque $\tau = F \times (\text{lever arm}) \times 2 = (ILB\sin\theta)(w/2) \times 2 = IAB\sin\theta$ for one turn. For $N$ turns: $\tau = NIAB\sin\theta$. (ii) $\mu = NIA = 100 \times 2.5 \times (0.04 \times 0.06) = 100 \times 2.5 \times 0.0024 = 0.60\ \text{A·m}^{2}$. (iii) $\tau_{\max} = NIAB\sin(90^{\circ}) = 100 \times 2.5 \times 0.0024 \times 0.40 = 0.24\ \text{N·m}$. Occurs when the plane of the coil is parallel to $\vec{B}$ (i.e., normal $\perp \vec{B}$).

Part (b): (i) $\tau(t) = NIAB\sin(\omega t) = 0.24\sin(120\pi t)\ \text{N·m}$. (ii) $\tau_{\text{avg}} = \frac{2}{\pi} NIAB = \frac{2}{\pi} \times 0.24 \approx 0.153\ \text{N·m}$. (iii) In an AC motor, the current alternates direction each half-cycle. When the current reverses, it does so at the same time the coil passes through the neutral plane, so the torque direction remains the same throughout the rotation.

Section C — C3 Transformer

Part (a): (i) $B = \mu_{0} n I = \mu_{0} (N/l) I$. Flux through one turn: $\Phi_{B} = BA = \mu_{0} NIA/l$. Flux linkage: $N\Phi_{B} = \mu_{0} N^{2} IA/l$. So $L = N\Phi_{B}/I = \mu_{0} N^{2} A/l$. (ii) $L = (4\pi\times10^{-7})(500)^{2}(4.0\times10^{-4})/(0.30) = (4\pi\times10^{-7})(2.5\times10^{5})(4.0\times10^{-4})/0.30 = (4\pi\times10^{-7})(100)/0.30 = (1.257\times10^{-4})/0.30 = 4.19\times10^{-4}\ \text{H} \approx 0.419\ \text{mH}$. (iii) $U = \frac{1}{2}LI^{2} = \frac{1}{2}(4.19\times10^{-4})(3.0)^{2} = \frac{1}{2}(4.19\times10^{-4})(9) = 1.89\times10^{-3}\ \text{J} \approx 1.89\ \text{mJ}$.

Part (b): (i) $M = k\sqrt{L_{1}L_{2}} = 0.80\sqrt{5.0\times10^{-3} \times 20\times10^{-3}} = 0.80\sqrt{1.0\times10^{-4}} = 0.80 \times 0.01 = 8.0\times10^{-3}\ \text{H} = 8.0\ \text{mH}$. (ii) $\mathcal{E}{2} = -M dI{1}/dt = -(8.0\times10^{-3})(200) = -1.6\ \text{V}$. Magnitude: $1.6\ \text{V}$. (iii) Reciprocity: $M_{12} = M_{21} = M$ — the mutual inductance from coil 1 to coil 2 equals that from coil 2 to coil 1.

Part (c): (i) Step-up ($N_{s} > N_{p}$). (ii) $V_{s} = V_{p} \times N_{s}/N_{p} = 240 \times 800/200 = 240 \times 4 = 960\ \text{V}$. $I_{s} = I_{p} \times N_{p}/N_{s} = 0.50 \times 200/800 = 0.125\ \text{A}$. (iii) $R_{L} = V_{s}/I_{s} = 960/0.125 = 7680\ \Omega \approx 7.68\ \text{k}\Omega$.

Section C — C4 Semiconductor & Biasing

Part (a): (i) Sketch: Input is a full sine wave ($\pm 10\ \text{V}$). Output is only the positive half-cycles, starting at $V_{\text{in}} = 0.7\ \text{V}$ (diode threshold). (ii) The diode begins conducting when $V_{\text{in}} > 0.7\ \text{V}$. (iii) Peak output $= 10 - 0.7 = 9.3\ \text{V}$.

Part (b): (i) $V_{B} = 15 \times 12/(56+12) = 15 \times 12/68 \approx 2.65\ \text{V}$. (ii) $V_{E} = V_{B} - V_{BE} = 2.65 - 0.7 = 1.95\ \text{V}$. $I_{E} = V_{E}/R_{E} = 1.95/1000 = 1.95\ \text{mA}$. $I_{C} \approx I_{E} = 1.95\ \text{mA}$. (iii) Voltage divider bias fixes $V_{B}$ independently of $\beta$. So $V_{E}$, $I_{E}$, and $I_{C}$ are all nearly independent of $\beta$. In fixed bias, $I_{C} = \beta I_{B}$ varies directly with $\beta$, making it unstable.

Part (c): (i) KCL at transistor: current entering (emitter) = current leaving (base + collector). So $I_{E} = I_{B} + I_{C}$. (ii) $I_{C} = \beta I_{B} = 150 \times 20\ \mu\text{A} = 3000\ \mu\text{A} = 3.0\ \text{mA}$. $I_{E} = I_{B} + I_{C} = 0.02 + 3.0 = 3.02\ \text{mA}$. (iii) $I_{E} = I_{B} + I_{C} = I_{B} + \beta I_{B} = (\beta+1)I_{B}$.

Part (d): (i) $\beta = 100$: $I_{B} = (20 - 0.7) / (470\times10^{3} + (101)(1.5\times10^{3})) = 19.3 / (470\times10^{3} + 151.5\times10^{3}) = 19.3 / 621.5\times10^{3} \approx 31.1\ \mu\text{A}$. $I_{C} = 100 \times 31.1\ \mu\text{A} = 3.11\ \text{mA}$. $V_{CE} = 20 - (3.11\times10^{-3})(3.3\times10^{3} + 1.5\times10^{3}) = 20 - 3.11\times10^{-3} \times 4.8\times10^{3} = 20 - 14.93 \approx 5.07\ \text{V}$. (ii) $\beta = 200$: $I_{B} = 19.3 / (470\times10^{3} + (201)(1.5\times10^{3})) = 19.3 / (470\times10^{3} + 301.5\times10^{3}) = 19.3 / 771.5\times10^{3} \approx 25.0\ \mu\text{A}$. $I_{C} = 200 \times 25.0\ \mu\text{A} = 5.00\ \text{mA}$. (iii) Percentage change in $I_{C} = (5.00 - 3.11)/3.11 \times 100% \approx 60.8%$. This shows emitter-stabilized bias provides moderate stability — $I_{C}$ still changes significantly with $\beta$, though less than fixed bias would (which would double). Voltage divider bias would be even more stable.

Section C — C5 Atomic Physics

Part (a): (i) Bohr's postulates: (1) Electrons orbit in circular orbits with Coulomb force providing centripetal acceleration. (2) Angular momentum is quantized: $mvr = n\hbar$. (3) Electrons emit/absorb photons of energy $hf = E_{i} - E_{f}$ when jumping between orbits. (ii) Derivation steps:

Coulomb $=$ centripetal: $\frac{1}{4\pi\varepsilon_{0}} \frac{e^{2}}{r_{n}^{2}} = \frac{m_{e}v_{n}^{2}}{r_{n}}$
Quantization: $m_{e}v_{n}r_{n} = n\hbar \Rightarrow v_{n} = n\hbar/(m_{e}r_{n})$
Substitute: $\frac{e^{2}}{4\pi\varepsilon_{0}r_{n}^{2}} = m_{e} \cdot \frac{n^{2}\hbar^{2}}{m_{e}^{2}r_{n}^{2}} \cdot \frac{1}{r_{n}}$
Simplify: $\frac{e^{2}}{4\pi\varepsilon_{0}} = \frac{n^{2}\hbar^{2}}{m_{e}r_{n}}$
Solve: $r_{n} = \frac{4\pi\varepsilon_{0}n^{2}\hbar^{2}}{m_{e}e^{2}} = n^{2}a_{0}$ (iii) $a_{0} = \frac{4\pi\varepsilon_{0}\hbar^{2}}{m_{e}e^{2}}$. $\hbar = h/(2\pi) = 6.63\times10^{-34}/(2\pi) = 1.055\times10^{-34}\ \text{J·s}$. $a_{0} = \frac{4\pi(8.85\times10^{-12})(1.055\times10^{-34})^{2}}{(9.11\times10^{-31})(1.60\times10^{-19})^{2}} = \frac{4\pi(8.85\times10^{-12})(1.113\times10^{-68})}{(9.11\times10^{-31})(2.56\times10^{-38})} = \frac{1.237\times10^{-78}}{2.332\times10^{-68}} \approx 5.29\times10^{-11}\ \text{m} = 0.529\ \text{Å}$.

Part (b): (i) $E_{1} = -13.6\ \text{eV}$, $E_{2} = -3.40\ \text{eV}$, $E_{3} = -1.51\ \text{eV}$, $E_{\infty} = 0$. (ii) $\Delta E = E_{3} - E_{2} = -1.51 - (-3.40) = 1.89\ \text{eV}$. $\lambda = hc/\Delta E = 1240/1.89 \approx 656\ \text{nm}$. (iii) Balmer series ($n \geq 3 \to n = 2$). Yes, $656\ \text{nm}$ is visible red light.

Section C — C6 Photoelectric Effect

Part (a): (i) $E = hc/\lambda = 1240/450 \approx 2.76\ \text{eV}$. In Joules: $E = 2.76 \times 1.60\times10^{-19} = 4.41\times10^{-19}\ \text{J}$. (ii) $f_{0} = \phi/h = (2.30 \times 1.60\times10^{-19})/(6.63\times10^{-34}) = (3.68\times10^{-19})/(6.63\times10^{-34}) \approx 5.55\times10^{14}\ \text{Hz}$. $\lambda_{c} = hc/\phi = 1240/2.30 \approx 539\ \text{nm}$. (iii) Yes, because $\lambda = 450\ \text{nm} < \lambda_{c} = 539\ \text{nm}$ (or $E = 2.76\ \text{eV} > \phi = 2.30\ \text{eV}$).

Part (b): (i) $K_{\text{max}} = hf - \phi = 2.76 - 2.30 = 0.46\ \text{eV}$. In Joules: $0.46 \times 1.60\times10^{-19} = 7.36\times10^{-20}\ \text{J}$. (ii) $V_{s} = K_{\text{max}}/e = 0.46\ \text{V}$. (iii) $v_{\text{max}} = \sqrt{2K_{\text{max}}/m_{e}} = \sqrt{2(7.36\times10^{-20})/(9.11\times10^{-31})} = \sqrt{1.615\times10^{11}} \approx 4.02\times10^{5}\ \text{m/s}$.

Part (c): (i) Graph: $K_{\text{max}}$ on $y$-axis vs $f$ on $x$-axis. Straight line with slope $= h$, $y$-intercept $= -\phi$, $x$-intercept $= f_{0} = \phi/h$. (ii) Doubling intensity at same frequency: (a) $K_{\text{max}}$ unchanged, (b) photocurrent doubles (more photons $\to$ more electrons). (iii) Larger $\phi$: line shifts right (larger $f_{0}$) and down (more negative $y$-intercept). Slope remains $h$.

Section D — Mixed & Reverse

$\tau = RC \Rightarrow C = \tau/R = 2.5/(50\times10^{3}) = 5.0\times10^{-5}\ \text{F} = 50\ \mu\text{F}$. $0.90 = 1 - e^{-t/2.5} \Rightarrow t = -2.5\ln(0.10) = 2.5 \times 2.303 \approx 5.76\ \text{s}$.
$B = \mu_{0} I r / (2\pi R^{2})$. Rearranging: $I = (2\pi R^{2} B)/(\mu_{0} r) = (2\pi)(0.002)^{2}(3.0\times10^{-4}) / ((4\pi\times10^{-7})(0.0005)) = (2\pi \times 4\times10^{-6} \times 3.0\times10^{-4}) / (4\pi\times10^{-7} \times 5\times10^{-4}) = (7.54\times10^{-9}) / (6.28\times10^{-10}) \approx 12.0\ \text{A}$.
$V_{s} = V_{p} \times N_{s}/N_{p} = 240 \times 1/10 = 24\ \text{V}$. $I_{s} = V_{s}/R_{L} = 24/12 = 2.0\ \text{A}$. $I_{p} = I_{s} \times N_{s}/N_{p} = 2.0 \times 1/10 = 0.20\ \text{A}$. $P = V_{s}I_{s} = 24 \times 2.0 = 48\ \text{W}$.
$V_{B} = 12 \times 10/(33+10) = 12 \times 10/43 \approx 2.79\ \text{V}$. $V_{E} = 2.79 - 0.7 = 2.09\ \text{V}$. $I_{E} = 2.09/1000 = 2.09\ \text{mA}$. $I_{C} \approx I_{E} = 2.09\ \text{mA}$. $V_{CE} = 12 - (2.09\times10^{-3})(2.0\times10^{3} + 1.0\times10^{3}) = 12 - 6.27 = 5.73\ \text{V}$. For $\beta = 240$, $I_{C}$ is nearly unchanged because voltage divider fixes $V_{B}$ (and hence $I_{C} \approx I_{E}$) independently of $\beta$. The Q-point is stable.
(i) $E = hc/\lambda = 1240/91.2 \approx 13.6\ \text{eV}$.
(ii) Yes, the atom will be ionized because $E_{\text{photon}} = 13.6\ \text{eV}$ equals the ionization energy ($E_{\infty} - E_{1} = 0 - (-13.6) = 13.6\ \text{eV}$).
(iii) $K_{\text{max}} = E_{\text{photon}} - (E_{\infty} - E_{1}) = 13.6 - 13.6 = 0\ \text{eV}$. The electron is just barely freed (zero kinetic energy at infinity).

Parting Exam Tips

Section A: Do all 15 questions in 30 min max. If stuck >3 min, move on.
Section B: Pick B1 + B2 + B3 (most predictable). Skip B4 unless you love AC theory.
Section C: Recommended selection: C4 + C3 + C5 + C1 (most predictable from leak).
Time: $A(30) + B(55) + C(75) = 160\ \text{min}$ total.
Never leave a question blank. Partial marks are better than zero.
Constants: Write all constants on the question paper immediately when the exam starts.