MetadataPastPaper.sampleTitle

First, calculate the number of pixels. Number of pixels along the width = \(8.0 \times 10^{-3}\text{ m} / (5.0 \times 10^{-6}\text{ m}) = 1600\). Number of pixels along the height = \(6.0 \times 10^{-3}\text{ m} / (5.0 \times 10^{-6}\text{ m}) = 1200\). Total number of pixels = \(1600 \times 1200 = 1.92 \times 10^6\). Since each pixel requires 12 bits of information, the total number of bits is \(1.92 \times 10^6 \times 12 = 2.304 \times 10^7\text{ bits}\). Converting this to bytes gives \(2.304 \times 10^7 / 8 = 2.88 \times 10^6\text{ bytes} = 2.88\text{ MB}\).

1 mark for the correct answer C. Method: Award mark for correct choice.

The total power \(P\) of thin lenses in close contact is the sum of their individual powers: \(P = P_1 + P_2 = +5.0\text{ D} + (-2.0\text{ D}) = +3.0\text{ D}\). The focal length \(f\) is the reciprocal of the power: \(f = 1 / P = 1 / 3.0\text{ D} \approx 0.33\text{ m} = 33\text{ cm}\).

1 mark for the correct answer B. Method: Award mark for correct choice.

Using the potential divider equation, \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{thermistor}} + R_{\text{fixed}}}\). Substituting the given values: \(V_{\text{out}} = 6.0\text{ V} \times \frac{2.0\text{ k}\Omega}{8.0\text{ k}\Omega + 2.0\text{ k}\Omega} = 6.0 \times \frac{2.0}{10.0} = 1.2\text{ V}\).

1 mark for the correct answer A. Method: Award mark for correct choice.

Resistivity \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). The percentage uncertainty in resistivity is given by the sum of the percentage uncertainties of the terms: \(\%\Delta \rho = \%\Delta R + 2 \times \%\Delta d + \%\Delta L\). First, find individual percentage uncertainties: \(\%\Delta R = (0.05 / 2.50) \times 100\% = 2.0\%\), \(\%\Delta L = (0.002 / 1.000) \times 100\% = 0.2\%\), \(\%\Delta d = (0.02 / 0.40) \times 100\% = 5.0\%\). Thus, \(\%\Delta \rho = 2.0\% + 2 \times 5.0\% + 0.2\% = 12.2\%\).

1 mark for the correct answer C. Method: Award mark for correct choice.

First, determine the extension \(\Delta x\) using the relationship \(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A \Delta x}\), which rearranges to \(\Delta x = \frac{F L}{A E} = \frac{300 \times 2.0}{1.5 \times 10^{-6} \times 2.0 \times 10^{11}} = 2.0 \times 10^{-3}\text{ m}\). The elastic strain energy stored is given by \(E_{\text{elastic}} = \frac{1}{2} F \Delta x = 0.5 \times 300 \times 2.0 \times 10^{-3} = 0.30\text{ J}\).

1 mark for the correct answer B. Method: Award mark for correct choice.

The horizontal component of velocity \(v_x\) remains constant at \(15\text{ m s}^{-1}\). The vertical component of velocity \(v_y\) is found using \(v_y^2 = u_y^2 + 2 g y\), where \(u_y = 0\text{ m s}^{-1}\), so \(v_y^2 = 2 \times 9.8 \times 40 = 784\text{ m}^2\text{ s}^{-2}\). The magnitude of the resultant velocity \(v\) is \(\sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 784} = \sqrt{225 + 784} = \sqrt{1009} \approx 31.8\text{ m s}^{-1}\), which rounds to \(32\text{ m s}^{-1}\) to two significant figures.

1 mark for the correct answer C. Method: Award mark for correct choice.

Impulse is given by the area under the force-time graph. The area can be divided into two parts: a triangle representing the first \(2.0\text{ s}\) and a rectangle representing the next \(3.0\text{ s}\). Area of the triangle = \(0.5 \times 2.0\text{ s} \times 10\text{ N} = 10\text{ N s}\). Area of the rectangle = \(3.0\text{ s} \times 10\text{ N} = 30\text{ N s}\). Total impulse = \(10\text{ N s} + 30\text{ N s} = 40\text{ N s}\).

1 mark for the correct answer B. Method: Award mark for correct choice.

The energy of an incident photon is \(E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{2.5 \times 10^{-7}} = 7.956 \times 10^{-19}\text{ J}\). Convert this photon energy to electronvolts: \(E = \frac{7.956 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 4.97\text{ eV}\). Using the photoelectric equation, the maximum kinetic energy is \(E_{\text{k,max}} = E - \Phi = 4.97\text{ eV} - 3.40\text{ eV} = 1.57\text{ eV}\), which rounds to \(1.6\text{ eV}\).

1 mark for the correct answer A. Method: Award mark for correct choice.

First, calculate the linear magnification \(m\):
\(m = \frac{\text{image width}}{\text{object width}} = \frac{0.036\text{ m}}{1.2\text{ m}} = 0.03\)

Since \(m = \frac{v}{u}\) (where \(v\) is image distance and \(u\) is object distance), we have:
\(v = 0.03u\)

Now apply the thin lens equation in terms of distances for a real image:
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)
\(\frac{1}{0.050} = \frac{1}{u} + \frac{1}{0.03u}\)
\(20 = \frac{1}{u} \left(1 + \frac{100}{3}\right)\)
\(20 = \frac{1}{u} \left(\frac{103}{3}\right)\)
\(3u = \frac{103}{20}\)
\(u = 1.72\text{ m} \approx 1.7\text{ m}\)

1 mark for the correct answer A.

Incorrect options represent:
- B: Errors in setting up magnification.
- C: Incorrect thin lens formula signs.
- D: Significant calculation errors in magnification ratios.

The potential divider equation for the output voltage across the thermistor is:
\(V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{therm}}}{R_{\text{therm}} + R_{\text{fixed}}}\)

Substitute the given values:
\(3.0\text{ V} = 9.0\text{ V} \times \frac{R_{\text{therm}}}{R_{\text{therm}} + 4.7\text{ k}\Omega}\)

Divide both sides by 9.0 V:
\(\frac{1}{3} = \frac{R_{\text{therm}}}{R_{\text{therm}} + 4.7\text{ k}\Omega}\)

Cross-multiply and solve for \(R_{\text{therm}}\):
\(R_{\text{therm}} + 4.7\text{ k}\Omega = 3R_{\text{therm}}\)
\(2R_{\text{therm}} = 4.7\text{ k}\Omega\)
\(R_{\text{therm}} = 2.35\text{ k}\Omega \approx 2.4\text{ k}\Omega\)

1 mark for the correct answer B.

Incorrect options represent:
- A: Arithmetic error during cross-multiplication.
- C: Matching the value of the fixed resistor instead of calculating.
- D: Incorrect formulation of the potential divider equation.

For the equation \(\rho = \frac{R \pi d^2}{4 L}\), the percentage uncertainty in \(\rho\) is determined by adding the percentage uncertainties of the components, remembering to multiply the uncertainty of \(d\) by 2 because it is squared:

\(\% \Delta \rho = \% \Delta R + 2 \times \% \Delta d + \% \Delta L\)
\(\% \Delta \rho = 2\% + 2(1.5\%) + 1.0\% = 2\% + 3\% + 1\% = 6.0\%\)

1 mark for the correct answer C.

Incorrect options represent:
- A: Forgetting to double the percentage uncertainty of the diameter (i.e. \(2\% + 1.5\% + 1.0\% = 4.5\%\)).
- B: Adding uncertainty of diameter once but with another arithmetic error.
- D: Incorrectly multiplying more terms or doubling other variables.

First, calculate the spring constant \(k\) of the wire:
\(k = \frac{E A}{L} = \frac{1.2 \times 10^{11}\text{ Pa} \times 1.0 \times 10^{-7}\text{ m}^2}{3.0\text{ m}} = 4.0 \times 10^3\text{ N m}^{-1}\)

Now, relate the elastic energy stored \(E_e\) to the extension \(x\):
\(E_e = \frac{1}{2} k x^2\)
\(0.20\text{ J} = \frac{1}{2} \times (4.0 \times 10^3\text{ N m}^{-1}) \times x^2\)
\(0.20 = 2000 x^2\)
\(x^2 = 1.0 \times 10^{-4}\text{ m}^2\)
\(x = 1.0 \times 10^{-2}\text{ m} = 10\text{ mm}\)

1 mark for the correct answer C.

Incorrect options represent:
- A: Calculation error omitting the factor of \(1/2\) in the energy formula.
- B: Incorrect division and factor-of-10 error.
- D: Forgetting to take the square root of \(x^2\) properly.

First, resolve the initial velocity into its vertical component \(u_y\):
\(u_y = u \sin \theta = 25\text{ m s}^{-1} \times \sin(30^\circ) = 12.5\text{ m s}^{-1}\)

At the maximum height, the vertical velocity component \(v_y = 0\).
Using the SUVAT equation \(v_y = u_y - g t\):
\(0 = 12.5 - 9.8 t\)
\(t = \frac{12.5}{9.8} \approx 1.28\text{ s} \approx 1.3\text{ s}\)

1 mark for the correct answer B.

Incorrect options represent:
- A: Dividing the vertical component incorrectly by another factor.
- C: Using \(\cos(30^\circ)\) instead of \(\sin(30^\circ)\).
- D: The total time of flight (twice the time to reach maximum height).

Calculate the energy of the incident photon in Joules:
\(E = hf = 6.63 \times 10^{-34}\text{ J s} \times 7.5 \times 10^{14}\text{ Hz} = 4.9725 \times 10^{-19}\text{ J}\)

Convert the work function \(\phi\) from electron-volts to Joules:
\(\phi = 2.1\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.36 \times 10^{-19}\text{ J}\)

Use Einstein's photoelectric equation to find the maximum kinetic energy:
\(E_{k,\text{max}} = hf - \phi = 4.9725 \times 10^{-19}\text{ J} - 3.36 \times 10^{-19}\text{ J} = 1.6125 \times 10^{-19}\text{ J} \approx 1.6 \times 10^{-19}\text{ J}\)

1 mark for the correct answer A.

Incorrect options represent:
- B: The work function value in Joules.
- C: The total incident photon energy in Joules.
- D: Adding the photon energy and work function together instead of subtracting.

First, find the grating spacing \(d\):
\(d = \frac{10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m}\)

The equation for diffraction maxima is:
\(d \sin \theta = n \lambda\)

The maximum possible value for \(\sin \theta\) is 1. Thus:
\(n \le \frac{d}{\lambda}\)
\(n \le \frac{2.0 \times 10^{-6}\text{ m}}{600 \times 10^{-9}\text{ m}} = 3.33\)

Since \(n\) must be an integer, the highest order maximum that can be observed is 3.

1 mark for the correct answer B.

Incorrect options represent:
- A: Underestimating the maximum integer value.
- C: Rounding 3.33 up to 4, which is physically impossible as \(\sin\theta > 1\).
- D: Mistake in calculating the grating spacing \(d\) by a factor of 10.

First, calculate the useful work done in raising the crate:
\(W_{\text{out}} = m g h = 80\text{ kg} \times 9.8\text{ m s}^{-2} \times 15\text{ m} = 11760\text{ J}\)

Next, find the useful power output:
\(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{11760\text{ J}}{6.0\text{ s}} = 1960\text{ W} = 1.96\text{ kW}\)

Now, calculate the efficiency \(\eta\):
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{1.96\text{ kW}}{2.5\text{ kW}} = 0.784 = 78.4\% \approx 78\%\)

1 mark for the correct answer C.

Incorrect options represent:
- A: Calculated using incorrect units or time intervals.
- B: Omitting the mass or gravity factor in power calculation.
- D: Dividing power input by power output (greater than 100% calculation error).

The potential difference across the thermistor is \(V_{\text{out}} = 2.4\text{ V}\). Since the total supply voltage is \(6.0\text{ V}\), the potential difference across the fixed resistor is \(V_R = 6.0\text{ V} - 2.4\text{ V} = 3.6\text{ V}\).

Using the ratio of voltages to resistances in a series loop:

\(\frac{V_{\text{th}}}{V_R} = \frac{R_{\text{th}}}{R}\)

\(\frac{2.4\text{ V}}{3.6\text{ V}} = \frac{R_{\text{th}}}{4.0\text{ k}\Omega}\)

\(R_{\text{th}} = \frac{2.4}{3.6} \times 4.0\text{ k}\Omega = 2.67\text{ k}\Omega \approx 2.7\text{ k}\Omega\).

C1: Reconstruct the potential divider formula or use current calculation to find the ratio. [1 mark]
A1: Correct resistance of 2.7 k\(\Omega\) selected (B). [1 mark]

1. Calculate photon energy:
\(E = hf = 6.63 \times 10^{-34}\text{ J s} \times 1.2 \times 10^{15}\text{ Hz} = 7.956 \times 10^{-19}\text{ J}\).

2. Convert the work function from eV to Joules:
\(\phi = 3.2\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 5.12 \times 10^{-19}\text{ J}\).

3. Apply Einstein's photoelectric equation:
\(E_{\text{k, max}} = E - \phi = 7.956 \times 10^{-19}\text{ J} - 5.12 \times 10^{-19}\text{ J} = 2.836 \times 10^{-19}\text{ J} \approx 2.8 \times 10^{-19}\text{ J}\).

C1: Convert work function to Joules or photon energy to eV. [1 mark]
A1: Correct answer A selected. [1 mark]

1. Use the Young modulus formula to find extension:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}\), so \(\Delta L = \frac{F L}{A E}\).

\(\Delta L = \frac{30\text{ N} \times 2.5\text{ m}}{1.5 \times 10^{-7}\text{ m}^2 \times 1.2 \times 10^{11}\text{ Pa}} = \frac{75}{18000} = 4.17 \times 10^{-3}\text{ m}\).

2. Calculate the strain energy (elastic potential energy) stored:
\(U = \frac{1}{2} F \Delta L = \frac{1}{2} \times 30\text{ N} \times 4.17 \times 10^{-3}\text{ m} = 0.0625\text{ J} \approx 6.3 \times 10^{-2}\text{ J}\).

C1: Calculate the extension of the wire first. [1 mark]
A1: Correct option B chosen. [1 mark]

1. Calculate the curvature of the wavefronts entering the lens from the object at distance \(u = -0.50\text{ m}\):

\(U = \frac{1}{u} = \frac{1}{-0.50\text{ m}} = -2.0\text{ D}\).

2. Apply the lens equation to find the curvature of the wavefronts leaving the lens:

\(V = U + P = -2.0\text{ D} + 6.0\text{ D} = +4.0\text{ D}\).

3. Calculate the image distance \(v\) from the curvature \(V\):

\(v = \frac{1}{V} = \frac{1}{+4.0\text{ D}} = +0.25\text{ m} = 25\text{ cm}\).

C1: Correctly calculate the initial curvature U and resultant curvature V. [1 mark]
A1: Correct option C chosen. [1 mark]

(a) Total number of pixels = \(2048 \times 2048 = 4,194,304\).
Total bits = \(4,194,304 \times 12\text{ bits} = 50,331,648\text{ bits}\).
Total bytes = \(50,331,648 / 8 = 6,291,456\text{ bytes}\).
File size in MB = \(6,291,456 / 10^6 = 6.29\text{ MB}\) (to 3 s.f.).

(b) Using the small angle approximation, the distance on the sensor \(s = f \theta\).
\(s = 1.50\text{ m} \times 4.00 \times 10^{-5}\text{ rad} = 6.00 \times 10^{-5}\text{ m}\) (or \(60.0\ \mu\text{m}\)).

(c) The separation between the star images in terms of pixels is:
\(60.0\ \mu\text{m} / 9.00\ \mu\text{m} \approx 6.67\text{ pixels}\).
Since the distance is significantly greater than 1 pixel (typically at least 2 pixels are needed to clearly resolve two peaks with a dark pixel in between), the stars will easily be resolved as separate objects.

Part (a): [3 marks total]
- Award 1 mark for calculating total pixels (\(4.19 \times 10^6\)) or total bits (\(5.03 \times 10^7\)).
- Award 1 mark for dividing total bits by 8 to get bytes (\(6.29 \times 10^6\)).
- Award 1 mark for correct final value of \(6.29\text{ MB}\) (accept 6.00 MB if using binary prefix factor of \(2^{20}\)).

Part (b): [2 marks total]
- Award 1 mark for selecting and using \(s = f\theta\).
- Award 1 mark for correct final answer: \(6.00 \times 10^{-5}\text{ m}\) (or \(60.0\ \mu\text{m}\)).

Part (c): [3.33 marks total]
- Award 1 mark for calculating the separation in pixels (\(6.67\text{ pixels}\)).
- Award 1.33 marks for the logical comparison that this is greater than the resolution limit (minimum 1 or 2 pixels) and stating the conclusion clearly.

(a) Potential divider equation: \(V_{\text{out}} = V_{\text{supply}} \times \frac{R}{R + R_{\text{th}}}\).
\(V_{\text{out}} = 6.00\text{ V} \times \frac{1.20\text{ k}\Omega}{1.20\text{ k}\Omega + 2.80\text{ k}\Omega} = 6.00 \times \frac{1.20}{4.00} = 1.80\text{ V}\) (as shown).

(b) Rearranging the potential divider equation for the new state:
\(3.60 = 6.00 \times \frac{1.20}{1.20 + R_{\text{th}}}\)
\(1.20 + R_{\text{th}} = \frac{6.00 \times 1.20}{3.60} = 2.00\text{ k}\Omega\)
\(R_{\text{th}} = 2.00 - 1.20 = 0.80\text{ k}\Omega = 800\ \Omega\).

(c) As temperature increases, the resistance of the NTC thermistor decreases exponentially. The change in resistance per degree Celsius (\(dR_{\text{th}}/dT\)) is much smaller at high temperatures than at low temperatures. Consequently, the rate of change of the output potential difference with respect to temperature (sensitivity) decreases as the temperature rises.

Part (a): [2 marks total]
- Award 1 mark for correct formulation of potential divider equation.
- Award 1 mark for substituting values to obtain exactly \(1.80\text{ V}\).

Part (b): [3 marks total]
- Award 1 mark for substituting the new output voltage into the potential divider equation.
- Award 1 mark for algebraic rearrangement to find total resistance of \(2.00\text{ k}\Omega\).
- Award 1 mark for correct final thermistor resistance of \(800\ \Omega\) (or \(0.80\text{ k}\Omega\)).

Part (c): [3.33 marks total]
- Award 1.33 marks for explaining that NTC resistance decreases non-linearly/exponentially with temperature.
- Award 1 mark for noting that the change in resistance per unit temperature is smaller at higher temperatures.
- Award 1 mark for concluding that the voltage sensitivity of the sensor decreases over this range.

(a) Cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\).
Resistivity \(\rho = \frac{R A}{L} = \frac{4.50\ \Omega \times 1.134 \times 10^{-7}\text{ m}^2}{1.250\text{ m}} = 4.08 \times 10^{-7}\ \Omega\text{ m}\).

(b) Percentage uncertainties for each variable:
\(\%\Delta R = \frac{0.05}{4.50} \times 100\% = 1.11\%\)
\(\%\Delta L = \frac{0.002}{1.250} \times 100\% = 0.16\%\)
\(\%\Delta d = \frac{0.02}{0.38} \times 100\% = 5.26\%\)
Since \(A \propto d^2\), \(\%\Delta A = 2 \times \%\Delta d = 10.53\%\).
Total percentage uncertainty in \(\rho\) is \(\%\Delta \rho = \%\Delta R + \%\Delta L + \%\Delta A = 1.11\% + 0.16\% + 10.53\% = 11.8\%\) (or \(12\%\) to 2 s.f.).
Absolute uncertainty = \(11.8\% \times 4.08 \times 10^{-7} = 0.48 \times 10^{-7}\ \Omega\text{ m}\) (or \(0.5 \times 10^{-7}\ \Omega\text{ m}\)).

(c) The diameter \(d\) contributes the most uncertainty (its percentage uncertainty is squared in the area calculation, accounting for \(10.5\%\) of the total \(11.8\%\)).
Improvement: Use a micrometer screw gauge (precision of \(0.01\text{ mm}\)) instead of calipers, and take multiple diameter measurements at different orientations along the wire to compute a mean.

Part (a): [3 marks total]
- Award 1 mark for correct calculation of area \(A = 1.13 \times 10^{-7}\text{ m}^2\).
- Award 1 mark for using \(\rho = RA/L\).
- Award 1 mark for correct resistivity value with units: \(4.08 \times 10^{-7}\ \Omega\text{ m}\).

Part (b): [3.33 marks total]
- Award 1 mark for individual percentage uncertainties of \(R\) (1.11%) and \(L\) (0.16%).
- Award 1 mark for doubling the percentage uncertainty of \(d\) to find the uncertainty in \(A\) (10.53%).
- Award 1.33 marks for correct summation to get \(11.8\%\) (or \(12\%\)) and absolute uncertainty of \(0.5 \times 10^{-7}\ \Omega\text{ m}\).

Part (c): [2 marks total]
- Award 1 mark for correctly identifying diameter as the largest contributor.
- Award 1 mark for a valid improvement (using a micrometer screw gauge and averaging multiple readings).

(a) Area \(A = \frac{\pi d^2}{4} = \frac{\pi (1.75 \times 10^{-3}\text{ m})^2}{4} = 2.405 \times 10^{-6}\text{ m}^2\).
Stress \(\sigma = \frac{F}{A} = \frac{45.0\text{ N}}{2.405 \times 10^{-6}\text{ m}^2} = 1.87 \times 10^7\text{ Pa}\) (or \(18.7\text{ MPa}\)).
Strain \(\epsilon = \frac{\Delta L}{L} = \frac{3.50 \times 10^{-3}\text{ m}}{2.00\text{ m}} = 1.75 \times 10^{-3}\) (or \(0.175\%\)).

(b) Young Modulus \(E = \frac{\sigma}{\epsilon} = \frac{1.871 \times 10^7\text{ Pa}}{1.75 \times 10^{-3}} = 1.07 \times 10^{10}\text{ Pa}\) (or \(10.7\text{ GPa}\)).

(c) Polymer elastic behavior: Polymers consist of long, tangled polymer chains. Under initial tension, these chains uncoil and straighten, which requires relatively little force and is fully reversible once the force is removed.
Contrast with metals: In metals, elastic deformation involves slightly increasing the separation between metal cations in the crystal lattice against electrostatic forces (stretching metallic bonds). This requires much higher forces, resulting in a much higher Young modulus than polymers.

Part (a): [3 marks total]
- Award 1 mark for correct cross-sectional area calculation.
- Award 1 mark for correct tensile stress with unit (\(1.87 \times 10^7\text{ Pa}\)).
- Award 1 mark for correct dimensionless tensile strain (\(1.75 \times 10^{-3}\)).

Part (b): [2 marks total]
- Award 1 mark for using \(E = \text{stress} / \text{strain}\).
- Award 1 mark for correct final Young modulus value (\(1.07 \times 10^{10}\text{ Pa}\) or \(10.7\text{ GPa}\)).

Part (c): [3.33 marks total]
- Award 1.33 marks for describing polymer elasticity as the uncoiling/unkinking of tangled long-chain molecules.
- Award 1 mark for describing metal elastic deformation as the stretching of electrostatic metallic bonds between lattice ions.
- Award 1 mark for comparing the relative difficulty/stiffness (noting metals are much stiffer/have a higher Young modulus).

(a) Horizontal component: \(u_x = 15.0 \cos(30.0^\circ) = 13.0\text{ m s}^{-1}\).
Vertical component: \(u_y = 15.0 \sin(30.0^\circ) = 7.50\text{ m s}^{-1}\).

(b) Define upwards as positive. The displacement \(s = -35.0\text{ m}\), \(u = 7.50\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\).
Using \(s = ut + \frac{1}{2}at^2\):
\(-35.0 = 7.50t - 4.905t^2\)
Rearranging into a quadratic equation:
\(4.905t^2 - 7.50t - 35.0 = 0\)
Using the quadratic formula:
\(t = \frac{7.50 \pm \sqrt{(-7.50)^2 - 4(4.905)(-35.0)}}{2(4.905)}\)
\(t = \frac{7.50 \pm \sqrt{56.25 + 686.7}}{9.81} = \frac{7.50 + 27.26}{9.81} = 3.54\text{ s}\) (discarding the negative time solution).

(c) The horizontal motion has constant velocity because air resistance is negligible.
\(x = u_x \times t = 12.99\text{ m s}^{-1} \times 3.543\text{ s} = 46.0\text{ m}\).

Part (a): [2 marks total]
- Award 1 mark for correct horizontal component: \(13.0\text{ m s}^{-1}\).
- Award 1 mark for correct vertical component: \(7.50\text{ m s}^{-1}\).

Part (b): [3.33 marks total]
- Award 1 mark for correct substitution into kinematic equation with consistent signs (e.g. \(-35.0 = 7.50t - 4.905t^2\)).
- Award 1.33 marks for setting up and solving the quadratic equation correctly.
- Award 1 mark for the correct final positive time of \(3.54\text{ s}\).

Part (c): [3 marks total]
- Award 1 mark for recognizing that horizontal velocity is constant.
- Award 1 mark for substituting the calculated time and horizontal velocity component.
- Award 1 mark for correct range of \(46.0\text{ m}\) (accept range \(45.5 - 46.5\text{ m}\) depending on rounding of intermediate steps).

(a) Photon energy in Joules:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} = 5.234 \times 10^{-19}\text{ J}\).
Converting to electronvolts:
\(E_{\text{eV}} = \frac{5.234 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 3.271\text{ eV} \approx 3.27\text{ eV}\).

(b) According to Einstein's photoelectric equation:
\(E_{k,\max} = hf - \Phi = 3.271\text{ eV} - 2.28\text{ eV} = 0.991\text{ eV}\).
Converting maximum kinetic energy back to Joules:
\(E_{k,\max} = 0.991 \times 1.60 \times 10^{-19}\text{ J} = 1.586 \times 10^{-19}\text{ J}\) (or \(1.59 \times 10^{-19}\text{ J}\)).

(c) Maximum kinetic energy remains unchanged because the photon energy (determined by the frequency/wavelength of light) remains the same.
The rate of emission of photoelectrons doubles because light intensity is proportional to the number of photons arriving per second; doubling the intensity doubles the number of incident photons, and each photon can release one photoelectron.

Part (a): [3 marks total]
- Award 1 mark for correct formula \(E = hc/\lambda\) and substitution.
- Award 1 mark for correct energy in Joules (\(5.23 \times 10^{-19}\text{ J}\)).
- Award 1 mark for correct division by \(1.60 \times 10^{-19}\) to get \(3.27\text{ eV}\).

Part (b): [2 marks total]
- Award 1 mark for using \(E_{k,\max} = hf - \Phi\) in consistent units.
- Award 1 mark for correct final value in Joules (\(1.58 \times 10^{-19}\text{ J}\) to \(1.60 \times 10^{-19}\text{ J}\)).

Part (c): [3.33 marks total]
- Award 1.33 marks for explaining that maximum kinetic energy depends only on photon energy and work function, hence it is unchanged.
- Award 1 mark for stating that doubling intensity means doubling the rate of arriving photons.
- Award 1 mark for concluding that the rate of photoelectron emission doubles due to the 1-to-1 interaction.

(a) The angular separation \( \theta \) corresponding to one pixel width is calculated using the small angle approximation:
\( \theta = \frac{\text{pixel width}}{\text{focal length}} = \frac{6.0 \times 10^{-6}\,\text{m}}{1.5\,\text{m}} = 4.0 \times 10^{-6}\,\text{rad} \).

(b) Comparing the angular separation of the binary star system with the minimum angular resolution:
\( 7.5 \times 10^{-6}\,\text{rad} > 4.0 \times 10^{-6}\,\text{rad} \).
Since the stars' separation is greater than the minimum pixel resolution, they can be resolved (the images will fall on non-adjacent pixels).

To find the physical separation \( s \) on the CCD:
\( s = f \theta = 1.5\,\text{m} \times 7.5 \times 10^{-6}\,\text{rad} = 1.125 \times 10^{-5}\,\text{m} \approx 1.1 \times 10^{-5}\,\text{m} \) (or \( 11\,\mu\text{m} \) to 2 significant figures).

Part (a):
- [1 mark] Correctly recalls or uses \( \theta = \frac{x}{f} \).
- [1 mark] Evaluates \( 4.0 \times 10^{-6}\,\text{rad} \) (ignore unit omission here but expect correct magnitude).

Part (b):
- [1 mark] Compares \( 7.5 \times 10^{-6}\,\text{rad} \) with the limit from (a) and concludes they can be resolved because the angle is larger.
- [1 mark] Uses \( s = f \theta \) with the correct values.
- [1 mark] Calculates image separation as \( 1.1 \times 10^{-5}\,\text{m} \) (accept \( 1.13 \times 10^{-5}\,\text{m} \) or \( 11\,\mu\text{m} \)).

(a) At \( 20.0^\circ\text{C} \), \( R_T = 1.80\,\text{k}\Omega \).
Using the potential divider formula:
\( V_{\text{out}} = V_{\text{in}} \times \frac{R_T}{R + R_T} \)
\( V_{\text{out}} = 5.00\,\text{V} \times \frac{1.80\,\text{k}\Omega}{1.20\,\text{k}\Omega + 1.80\,\text{k}\Omega} = 5.00 \times \frac{1.80}{3.00} = 3.00\,\text{V} \).

(b) First, calculate \( V_{\text{out}} \) at \( 30.0^\circ\text{C} \) where \( R_T = 1.30\,\text{k}\Omega \):
\( V_{\text{out}} = 5.00\,\text{V} \times \frac{1.30\,\text{k}\Omega}{1.20\,\text{k}\Omega + 1.30\,\text{k}\Omega} = 5.00 \times \frac{1.30}{2.50} = 2.60\,\text{V} \).

Next, find the change in output voltage \( \Delta V_{\text{out}} \):
\( \Delta V_{\text{out}} = 2.60\,\text{V} - 3.00\,\text{V} = -0.40\,\text{V} \).

The change in temperature \( \Delta T \):
\( \Delta T = 30.0^\circ\text{C} - 20.0^\circ\text{C} = 10.0^\circ\text{C} \).

Average sensitivity \( = \frac{\Delta V_{\text{out}}}{\Delta T} = \frac{-0.40\,\text{V}}{10.0^\circ\text{C}} = -0.040\,\text{V}^\circ\text{C}^{-1} \) (or \( -0.040\,\text{V}\,\text{K}^{-1} \)).

Part (a):
- [1 mark] Substitution of correct values into the potential divider equation.
- [1 mark] Evaluates \( 3.00\,\text{V} \).

Part (b):
- [1 mark] Calculates \( V_{\text{out}} \) at \( 30.0^\circ\text{C} \) as \( 2.60\,\text{V} \).
- [1 mark] Determines change in voltage of \( -0.40\,\text{V} \) (allow \( 0.40\,\text{V} \)) and divides by \( 10.0^\circ\text{C} \).
- [1 mark] Correct final sensitivity with correct unit (accept \( -0.040\,\text{V}^\circ\text{C}^{-1} \), \( -0.040\,\text{V}\,\text{K}^{-1} \) or positive equivalent magnitude if clear, but unit must be fully correct).

(a) Young modulus \( E \) is given by:
\( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L} \)
Substitute the given values:
\( F = 45.0\,\text{N} \)
\( L = 2.40\,\text{m} \)
\( A = 1.50 \times 10^{-7}\,\text{m}^2 \)
\( \Delta L = 3.60 \times 10^{-3}\,\text{m} \)

\( E = \frac{45.0 \times 2.40}{1.50 \times 10^{-7} \times 3.60 \times 10^{-3}} = \frac{108}{5.40 \times 10^{-10}} = 2.00 \times 10^{11}\,\text{Pa} \).

(b) The elastic strain energy \( W \) is given by:
\( W = \frac{1}{2} F \Delta L \)
\( W = 0.5 \times 45.0\,\text{N} \times 3.60 \times 10^{-3}\,\text{m} = 0.081\,\text{J} \) (or \( 8.1 \times 10^{-2}\,\text{J} \)).

Part (a):
- [1 mark] Calculates stress \( (3.00 \times 10^8\,\text{Pa}) \) OR strain \( (1.50 \times 10^{-3}) \).
- [1 mark] Rearranges and substitutes correctly into \( E = \frac{FL}{A\Delta L} \).
- [1 mark] Calculates \( 2.00 \times 10^{11}\,\text{Pa} \) (or \( \text{N}\,\text{m}^{-2} \)).

Part (b):
- [1 mark] Recalls and uses \( W = \frac{1}{2} F \Delta L \).
- [1 mark] Calculates \( 0.081\,\text{J} \) (or \( 8.1 \times 10^{-2}\,\text{J} \)).

(a) The vertical component of the initial velocity, \( u_y \), is found resolved vertically:
\( u_y = u \sin \theta = 8.50 \sin(30.0^\circ) = 4.25\,\text{m}\,\text{s}^{-1} \).

(b) Defining upwards as the positive direction:
\( u_y = +4.25\,\text{m}\,\text{s}^{-1} \)
\( s_y = -1.80\,\text{m} \) (since the ball ends up \( 1.80\,\text{m} \) below the launch point)
\( a_y = -9.81\,\text{m}\,\text{s}^{-2} \)

Using the kinematic equation:
\( s_y = u_y t + \frac{1}{2} a_y t^2 \)
\( -1.80 = 4.25 t - 4.905 t^2 \)

Rearranging into a quadratic equation:
\( 4.905 t^2 - 4.25 t - 1.80 = 0 \)

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( t = \frac{4.25 \pm \sqrt{(-4.25)^2 - 4(4.905)(-1.80)}}{2 \times 4.905} \)
\( t = \frac{4.25 \pm \sqrt{18.0625 + 35.316}}{9.81} \)
\( t = \frac{4.25 \pm \sqrt{53.3785}}{9.81} \)
\( t = \frac{4.25 \pm 7.306}{9.81} \)

Choosing the positive time root:
\( t = \frac{11.556}{9.81} \approx 1.18\,\text{s} \).

Part (a):
- [1 mark] Shows \( 8.50 \sin(30.0^\circ) = 4.25\,\text{m}\,\text{s}^{-1} \).

Part (b):
- [1 mark] Selects correct vertical kinematic equation \( s_y = u_y t + \frac{1}{2} a_y t^2 \).
- [1 mark] Correct substitution of signs (e.g., negative displacement and acceleration, positive initial vertical velocity).
- [1 mark] Solves quadratic equation or uses alternative two-stage method (e.g., finding time to peak then time to drop).
- [1 mark] Correct final time of flight of \( 1.18\,\text{s} \) (accept \( 1.2\,\text{s} \)).

(a) The energy of a photon \( E \) is given by:
\( E = \frac{h c}{\lambda} \)
Using standard constants \( h = 6.63 \times 10^{-34}\,\text{J}\,\text{s} \) and \( c = 3.00 \times 10^8\,\text{m}\,\text{s}^{-1} \):
\( E = \frac{6.63 \times 10^{-34}\,\text{J}\,\text{s} \times 3.00 \times 10^8\,\text{m}\,\text{s}^{-1}}{2.40 \times 10^{-7}\,\text{m}} \)
\( E = \frac{1.989 \times 10^{-25}}{2.40 \times 10^{-7}} = 8.2875 \times 10^{-19}\,\text{J} \approx 8.3 \times 10^{-19}\,\text{J} \).

(b) The work function \( \phi \) in joules is:
\( \phi = 4.30\,\text{eV} \times 1.60 \times 10^{-19}\,\text{J}\,\text{eV}^{-1} = 6.88 \times 10^{-19}\,\text{J} \).

By Einstein's photoelectric equation:
\( E_{k\text{,max}} = E_{\text{photon}} - \phi \)
\( E_{k\text{,max}} = 8.29 \times 10^{-19}\,\text{J} - 6.88 \times 10^{-19}\,\text{J} = 1.41 \times 10^{-19}\,\text{J} \).
(If the rounded value of \( 8.3 \times 10^{-19}\,\text{J} \) is used, \( E_{k\text{,max}} = 8.30 \times 10^{-19} - 6.88 \times 10^{-19} = 1.42 \times 10^{-19}\,\text{J} \). Both are acceptable).

Part (a):
- [1 mark] Correct substitution of \( h, c, \lambda \) into \( E = \frac{hc}{\lambda} \).
- [1 mark] Shows calculation resulting in at least 3 significant figures \( (8.29 \times 10^{-19}\,\text{J}) \) to justify rounding to \( 8.3 \times 10^{-19}\,\text{J} \).

Part (b):
- [1 mark] Converts work function to joules: \( 4.30 \times 1.60 \times 10^{-19} = 6.88 \times 10^{-19}\,\text{J} \).
- [1 mark] Subtracts work function from photon energy (either \( 8.29 \times 10^{-19}\,\text{J} \) or \( 8.30 \times 10^{-19}\,\text{J} \)).
- [1 mark] Correct final kinetic energy of \( 1.41 \times 10^{-19}\,\text{J} \) or \( 1.42 \times 10^{-19}\,\text{J} \).

(a) To make \(V_{out}\) increase with temperature, the fixed resistor \(R\) and the thermistor must be placed in series across the supply, with \(V_{out}\) measured across the fixed resistor \(R\). As temperature increases, the resistance of the NTC thermistor decreases. This reduces the total resistance of the circuit, which increases the current flowing through it. Since \(V_{out} = I \times R\), and \(R\) is constant, the output voltage across \(R\) must increase.

(b)(i) Using the potential divider equation:
\(V_{out} = \frac{R}{R + R_{T}} \times V_{in}\)
\(V_{out} = \frac{1.2\text{ k}\Omega}{1.2\text{ k}\Omega + 2.4\text{ k}\Omega} \times 6.0\text{ V} = \frac{1.2}{3.6} \times 6.0 = 2.0\text{ V}\).

(b)(ii) Given \(V_{out} = 3.6\text{ V}\):
\(3.6 = \frac{1.2}{1.2 + R_{T}} \times 6.0\)
\(\frac{3.6}{6.0} = 0.6 = \frac{1.2}{1.2 + R_{T}}\)
\(1.2 + R_{T} = \frac{1.2}{0.6} = 2.0\text{ k}\Omega\)
\(R_{T} = 2.0 - 1.2 = 0.80\text{ k}\Omega\) (or \(800\text{ }\Omega\)).

(c) Experimental procedure:
1. Place the NTC thermistor in a beaker of water (or water bath) alongside a high-quality thermometer/temperature probe.
2. Connect the thermistor into the potential divider circuit with the \(6.0\text{ V}\) supply and a digital voltmeter connected across the fixed resistor \(R\).
3. Heat the water slowly and record the temperature and the corresponding voltmeter reading at regular intervals (e.g., every \(5^\circ\text{C}\)) over the desired range (e.g., \(0^\circ\text{C}\) to \(80^\circ\text{C}\)).
4. Ensure accuracy by stirring the water constantly to maintain uniform temperature and ensuring the thermometer is placed close to the thermistor.
5. Repeat the experiment during cooling to check for reproducibility and take averages.
6. Plot a calibration curve of \(V_{out}\) (y-axis) against Temperature (x-axis).

(d) The resistance of an NTC thermistor decreases exponentially/non-linearly as temperature increases; the rate of change of resistance with temperature is much smaller at high temperatures. In the potential divider, as \(R_T \to 0\) at very high temperatures, the output voltage \(V_{out}\) asymptotically approaches the supply voltage (\(6.0\text{ V}\)). Consequently, the rate of change of \(V_{out}\) with temperature decreases significantly, causing the overall sensitivity of the sensor to drop to near zero at high temperatures.

(a) [3 marks]
- 1 mark: Identifies that \(V_{out}\) must be measured across the fixed resistor \(R\).
- 1 mark: States that the resistance of the NTC thermistor decreases as temperature increases.
- 1 mark: Connects this to an increase in current/proportion of voltage across \(R\) to explain why \(V_{out}\) increases.

(b)(i) [2 marks]
- 1 mark: Correct substitution into potential divider equation.
- 1 mark: Clearly shows steps leading to exactly \(2.0\text{ V}\).

(b)(ii) [2 marks]
- 1 mark: Rearranges equation to find total resistance or equivalent ratio (e.g., \(1.2 + R_T = 2.0\)).
- 1 mark: Obtains \(0.80\text{ k}\Omega\) (accept \(800\text{ }\Omega\)).

(c) [5 marks]
- 1 mark: Mentions using a water bath/beaker of water and a thermometer/temperature probe.
- 1 mark: Details connecting a voltmeter across the fixed resistor to measure \(V_{out}\).
- 1 mark: Explains taking readings at regular temperature intervals (e.g., every \(5\)-\(10^\circ\text{C}\)).
- 1 mark: Practical detail: stirring the water to ensure uniform temperature, or keeping probe and thermistor close.
- 1 mark: Identifies how to use the data (plot a calibration graph of \(V_{out}\) vs temperature).

(d) [3 marks]
- 1 mark: States that thermistor resistance change is non-linear/rate of decrease slows down at high temperatures.
- 1 mark: Explains that \(V_{out}\) approaches the limit of the supply voltage \(6.0\text{ V}\).
- 1 mark: Concludes that the sensitivity (\(\Delta V_{out} / \Delta T\)) of the circuit decreases at high temperatures.

(a) Measurements and instruments:
1. Original length (\(L\)) - measured using a metre rule / tape measure.
2. Diameter (\(d\)) - measured using a micrometer screw gauge (at multiple points and orientations).
3. Extension (\(x\)) - measured using a vernier scale / travelling microscope (or a pointer on a scale).
4. Applied force/load (\(F\)) - determined from the mass of suspended loads (masses \(\times g\)) or using a force meter.

(b)(i) Cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi (0.36 \times 10^{-3}\text{ m})^2}{4} = 1.018 \times 10^{-7}\text{ m}^2\).
Tensile stress \(\sigma = \frac{F}{A} = \frac{24.0\text{ N}}{1.018 \times 10^{-7}\text{ m}^2} = 2.358 \times 10^8\text{ Pa} \approx 2.36 \times 10^8\text{ Pa}\).

(b)(ii) Strain \(\epsilon = \frac{x}{L} = \frac{3.9 \times 10^{-3}\text{ m}}{2.00\text{ m}} = 1.95 \times 10^{-3}\).
Young modulus \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{2.358 \times 10^8\text{ Pa}}{1.95 \times 10^{-3}} = 1.21 \times 10^{11}\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(c) Uncertainty analysis:
- Percentage uncertainty in \(d\) = \(\frac{0.02}{0.36} \times 100\% = 5.56\%\).
- Since \(A \propto d^2\), percentage uncertainty in \(A\) = \(2 \times 5.56\% = 11.11\%\).
- Absolute uncertainty in \(A\) = \(11.11\% \times 1.018 \times 10^{-7}\text{ m}^2 = 1.13 \times 10^{-8}\text{ m}^2\) (or \(1.1 \times 10^{-8}\text{ m}^2\)).
- Percentage uncertainty in \(L\) = \(\frac{0.01}{2.00} \times 100\% = 0.50\%\).
- Percentage uncertainty in \(F\) = \(\frac{0.5}{24.0} \times 100\% = 2.08\%\).
- Percentage uncertainty in \(x\) = \(\frac{0.1}{3.9} \times 100\% = 2.56\%\).
- Adding percentage uncertainties (standard A-level approach):
Total \% uncertainty in \(E\) = \(\%\Delta F + \%\Delta L + \%\Delta A + \%\Delta x = 2.08\% + 0.50\% + 11.11\% + 2.56\% = 16.25\% \approx 16.3\%\) (accept \(16\%\)).
- (Alternative Quadrature method: \(\sqrt{2.08^2 + 0.50^2 + 11.11^2 + 2.56^2} \approx 11.6\%\)).

(d) Using a longer wire increases the magnitude of the extension for any given load (since \(x = \frac{FL}{AE}\)). Since the absolute uncertainty in the extension measurement remains constant (e.g. \(\pm 0.1\text{ mm}\)), a larger extension value significantly reduces the percentage uncertainty of the extension, leading to a more precise Young modulus.
Safety precaution: Safety goggles must be worn because the wire is under high tension and can snap violently, potentially causing eye injury; a catch-box containing sand or bubble wrap should be placed under the weights to cushion their fall.

(a) [3 marks]
- 1 mark: Mentions measuring diameter using a micrometer screw gauge.
- 1 mark: Mentions measuring original length using a metre rule / tape measure.
- 1 mark: Mentions measuring extension using a vernier scale / travelling microscope.

(b)(i) [2 marks]
- 1 mark: Correct calculation of cross-sectional area (\(1.02 \times 10^{-7}\text{ m}^2\)).
- 1 mark: Correct calculation of stress with correct unit (\(2.36 \times 10^8\text{ Pa}\) or \(\text{N m}^{-2}\)).

(b)(ii) [2 marks]
- 1 mark: Correct calculation of strain (\(1.95 \times 10^{-3}\)) or formula substitution \(E = FL/Ax\).
- 1 mark: Correct value for \(E\) (\(1.21 \times 10^{11}\text{ Pa}\) or \(1.2 \times 10^{11}\text{ Pa}\)).

(c) [5 marks]
- 1 mark: Calculates percentage uncertainty in diameter (\(5.56\%\)).
- 1 mark: Doubles the percentage uncertainty of diameter to get the area percentage uncertainty (\(11.1\%\)).
- 1 mark: Calculates absolute uncertainty in area (\(1.1 \times 10^{-8}\text{ m}^2\)).
- 1 mark: Calculates percentage uncertainties of other variables (\(F\) is \(2.1\%\), \(x\) is \(2.6\%\), \(L\) is \(0.5\%\)).
- 1 mark: Correctly sums or combines uncertainties in quadrature to get final percentage uncertainty (\(16.3\%\) or \(11.6\%\)).

(d) [3 marks]
- 1 mark: Connects longer wire to a larger extension (for the same stress).
- 1 mark: Explains that larger extension reduces percentage uncertainty of the extension measurement.
- 1 mark: Provides a valid safety precaution (e.g. safety goggles in case wire snaps, catch-box for falling masses).

(a) Photon model vs Wave theory:
- Threshold Frequency: In the photon model, light consists of packets of energy called photons (\(E = hf\)). A single electron can only absorb a single photon (one-to-one interaction). If the photon energy is less than the work function (\(\phi\)) of the metal, no electron can escape, regardless of intensity. Wave theory predicts energy builds up continuously over time, meaning any frequency should eventually cause emission if the wave is intense enough.
- Instantaneous Emission: In the photon model, energy transfer is instantaneous during the collision between a single photon and an electron. In wave theory, the wavefront is spread out over many atoms, so a low-intensity wave would require a measurable time delay for an individual electron to accumulate enough energy to escape.
- Intensity and Kinetic Energy: Increasing intensity means more photons per second, resulting in more photoelectrons emitted per second (higher current). However, the energy of individual photons (\(hf\)) remains unchanged, so the maximum kinetic energy of the emitted electrons (\(E_{k,max} = hf - \phi\)) remains unchanged. Wave theory predicts that more intense waves carry larger electric field amplitudes, which would deliver more energy to each electron, increasing their kinetic energy.

(b)(i) Energy of a photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{254 \times 10^{-9}\text{ m}} = 7.83 \times 10^{-19}\text{ J}\) (which is approximately \(7.8 \times 10^{-19}\text{ J}\)).

(b)(ii) The maximum kinetic energy of the emitted electrons in Joules is:
\(E_{k,max} = e V_s = 1.60 \times 10^{-19}\text{ C} \times 1.25\text{ V} = 2.00 \times 10^{-19}\text{ J}\) (or \(1.25\text{ eV}\)).
Using Einstein's photoelectric equation:
\(hf = \phi + E_{k,max}\)
\(\phi = hf - E_{k,max} = 7.83 \times 10^{-19}\text{ J} - 2.00 \times 10^{-19}\text{ J} = 5.83 \times 10^{-19}\text{ J}\).
To convert this to electron-volts:
\(\phi = \frac{5.83 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J/eV}} = 3.64\text{ eV}\).

(c)(i) Shorter wavelength (\(\lambda\)) means higher frequency (\(f = c/\lambda\)), so each incident photon has more energy. Since \(E_{k,max} = hf - \phi\) and the work function is constant, the maximum kinetic energy of the emitted photoelectrons increases.

(c)(ii) Intensity is power per unit area (\(I = N \times hf\) where \(N\) is the number of photons per unit area per second). Since intensity is kept constant but individual photon energy \(hf\) increases, the number of photons striking the surface per second must decrease. Because of the one-to-one interaction, fewer photoelectrons are emitted per second, so the maximum photoelectric current decreases.

(a) [6 marks] Level of Response:
- Detailed explanation of all three observations using the photon model AND clear contrast with classical wave theory predictions.
- Level 3 (5-6 marks): Comprehensive explanation of all 3 points, contrasting both models clearly with precise physics terminology.
- Level 2 (3-4 marks): Explains 2 points well, or all 3 points but lacks clear contrast with wave theory.
- Level 1 (1-2 marks): Basic recall of photon model or threshold frequency definition, but unstructured.

(b)(i) [2 marks]
- 1 mark: Correct equation \(E = hc/\lambda\) and substitution.
- 1 mark: Evaluation showing \(7.83 \times 10^{-19}\text{ J}\).

(b)(ii) [3 marks]
- 1 mark: Calculates \(E_{k,max} = 1.25\text{ eV}\) (or \(2.00 \times 10^{-19}\text{ J}\)).
- 1 mark: Uses Einstein's equation to find \(\phi = 5.83 \times 10^{-19}\text{ J}\).
- 1 mark: Correctly converts to \(3.64\text{ eV}\) (accept \(3.6\text{ eV}\) to \(3.7\text{ eV}\)).

(c)(i) [2 marks]
- 1 mark: Links shorter wavelength to higher photon energy.
- 1 mark: Concludes maximum kinetic energy increases (referring to Einstein's equation).

(c)(ii) [2 marks]
- 1 mark: Explains that constant intensity with higher photon energy means fewer photons strike the surface per second.
- 1 mark: Concludes there are fewer emitted electrons per second, leading to a decreased maximum current.