2023 IB DP Physics Practice Paper with Answers

Initial total kinetic energy is \(E_i = \frac{1}{2}m(2v)^2 + \frac{1}{2}(2m)v^2 = 2mv^2 + mv^2 = 3mv^2\). The total initial momentum is \(p_i = m(2v) + 2m(v) = 4mv\). After the inelastic collision, the combined mass is \(3m\), and its velocity is \(V = \frac{p_i}{3m} = \frac{4}{3}v\). The final kinetic energy is \(E_f = \frac{1}{2}(3m)V^2 = \frac{1}{2}(3m)\left(\frac{4}{3}v\right)^2 = \frac{8}{3}mv^2\). The kinetic energy lost is \(\Delta E = E_i - E_f = 3mv^2 - \frac{8}{3}mv^2 = \frac{1}{3}mv^2\). The fraction of initial kinetic energy lost is \(\frac{\Delta E}{E_i} = \frac{\frac{1}{3}mv^2}{3mv^2} = \frac{1}{9}\).

Award 1 mark for the correct option A. Method: Identify total initial kinetic energy and total initial momentum, apply conservation of momentum to find the final velocity, calculate the final kinetic energy, and then compute the fraction of energy lost.

At the air-film boundary, light reflects from a medium of lower refractive index (air, 1.0) to a higher one (film, \(n\)), resulting in a \(\pi\) phase change. At the film-glass boundary, light reflects from a medium of lower refractive index (film, \(n\)) to a higher one (glass, \(n_{\text{glass}} > n\)), also resulting in a \(\pi\) phase change. Because both reflections experience the same phase change, there is zero relative phase difference introduced by the reflections. Therefore, for destructive interference, the path difference within the film (which is \(2t\)) must equal a half-integer number of wavelengths in the medium: \(2t = (m + \frac{1}{2})\lambda_n = (m + \frac{1}{2})\frac{\lambda}{n}\). For the minimum non-zero thickness, we set \(m = 0\), which yields \(2t = \frac{\lambda}{2n}\), so \(t = \frac{\lambda}{4n}\).

Award 1 mark for the correct option A. Method: Deduce that both boundary reflections introduce a \(\pi\) phase shift, leading to no net relative phase change from reflection. Use the destructive interference condition for a thin film to solve for the minimum non-zero thickness.

According to Wien's displacement law, the wavelength of maximum intensity is inversely proportional to the absolute temperature: \(\lambda_{\max} \propto \frac{1}{T}\). Thus, doubling the temperature reduces the peak wavelength to \(\frac{\lambda_0}{2}\). According to the Stefan-Boltzmann law, the total power radiated is proportional to the fourth power of the absolute temperature: \(P \propto T^4\). Thus, doubling the temperature increases the power by a factor of \(2^4 = 16\), resulting in \(16 P_0\).

Award 1 mark for the correct option A. Method: Identify and apply Wien's displacement law for peak wavelength and the Stefan-Boltzmann law for radiated power.

The drift speed is given by the formula \(v = \frac{I}{nAe}\). Since both wires are made of copper, the number density of free electrons \(n\) is the same. The second wire has twice the diameter, which means its radius is also doubled, making its cross-sectional area \(A_2 = 4A\). The current in the second wire is \(4I\). Substituting these values into the drift speed formula gives \(v_2 = \frac{4I}{n(4A)e} = \frac{I}{nAe} = v\).

Award 1 mark for the correct option C. Method: Relate current to drift velocity and cross-sectional area. Determine how doubling the diameter affects the area, then evaluate the change in drift velocity.

During the first step (isothermal expansion), the temperature remains \(T\). According to Boyle's law, the pressure becomes \(p' = \frac{p V}{2V} = 0.5p\). In the second step, the gas is heated at constant pressure \(p' = 0.5p\) from volume \(2V\) to volume \(3V\). Applying Charles's law, the ratio of volume to temperature remains constant: \(\frac{2V}{T} = \frac{3V}{T_{\text{final}}}\), which gives \(T_{\text{final}} = 1.5 T\).

Award 1 mark for the correct option B. Method: Determine the intermediate state parameters after isothermal expansion, then apply Charles's law or the ideal gas equation to find the final temperature.

The energy of the emitted photon is equal to the difference in energy levels. For the first transition: \(\Delta E_1 = -1.0\text{ eV} - (-4.0\text{ eV}) = 3.0\text{ eV}\). Since \(\Delta E_1 = hf\), we have \(hf = 3.0\text{ eV}\). For the second transition: \(\Delta E_2 = -4.0\text{ eV} - (-9.0\text{ eV}) = 5.0\text{ eV}\). Since \(\Delta E_2 = hf_2\), we have \(hf_2 = 5.0\text{ eV}\). Dividing the two energy equations gives \(\frac{f_2}{f} = \frac{5.0}{3.0}\), so \(f_2 = \frac{5}{3} f\).

Award 1 mark for the correct option C. Method: Calculate the energy difference for both transitions, apply the photon energy equation \(E = hf\), and find the ratio of the frequencies.

For a pipe open at one end and closed at the other, only odd harmonics are present. The fundamental frequency (first harmonic) is \(f_1 = \frac{v}{4L}\). The third harmonic has a frequency of \(f_3 = 3 f_1 = \frac{3v}{4L}\). The difference between these two frequencies is \(\Delta f = f_3 - f_1 = \frac{3v}{4L} - \frac{v}{4L} = \frac{2v}{4L} = \frac{v}{2L}\).

Award 1 mark for the correct option B. Method: Recall the frequency formula for odd harmonics in a pipe with one closed end, determine the frequencies of the 1st and 3rd harmonics, and find their difference.

When the rod is horizontal, the gravitational force \(Mg\) acts at its center of mass, which is located at a distance of \(\frac{L}{2}\) from the pivot. The torque \(\tau\) acting on the rod is \(\tau = Mg \left(\frac{L}{2}\right) = \frac{1}{2}MgL\). Using Newton's second law for rotation, \(\tau = I\alpha\), we get \(\frac{1}{2}MgL = \left(\frac{1}{3}ML^2\right)\alpha\). Solving for \(\alpha\) gives \(\alpha = \frac{3g}{2L}\).

Award 1 mark for the correct option D. Method: Identify the position of the center of mass to compute the torque due to gravity, apply Newton's second law for rotation, and solve for the angular acceleration.

The total initial momentum of the system is \( p_{\text{initial}} = (m)(v) + (3m)(-\frac{v}{3}) = mv - mv = 0 \). By the conservation of linear momentum, the total final momentum must also be zero. Therefore, the common final velocity of the combined mass is zero.

Award [1] for the correct answer A.

For a single slit diffraction, the position of the first minimum is given by \( \sin \theta \approx \theta = \frac{\lambda}{b} \). If the new wavelength is \( \lambda' = \frac{\lambda}{2} \) and the new slit width is \( b' = 2b \), the new angle is \( \theta' = \frac{\lambda'}{b'} = \frac{\frac{\lambda}{2}}{2b} = \frac{\lambda}{4b} = \frac{\theta}{4} \).

Award [1] for the correct answer D.

Luminosity is given by \( L = \sigma A T^4 = 4\pi R^2 \sigma T^4 \). Thus, the ratio of luminosities is \( \frac{L_X}{L_Y} = \frac{R_X^2 T_X^4}{R_Y^2 T_Y^4} = \frac{R^2 T^4}{(\frac{R}{2})^2 (2T)^4} = \frac{R^2 T^4}{\frac{R^2}{4} \cdot 16 T^4} = \frac{1}{4} \).

Award [1] for the correct answer B.

The current is given by \( I = \frac{E}{R + r} \). As the external resistance \( R \) increases, the total resistance of the circuit increases, so the current \( I \) decreases. The terminal potential difference is given by \( V = E - Ir \). Since the current \( I \) decreases, the term \( Ir \) decreases, which means \( V \) increases.

Award [1] for the correct answer A.

The average random kinetic energy of ideal gas molecules is directly proportional to the absolute temperature \( T \) in Kelvin. First, convert the temperatures to Kelvin: \( T_{\text{initial}} = 27 + 273 = 300\text{ K} \) and \( T_{\text{final}} = 327 + 273 = 600\text{ K} \). The ratio is \( \frac{T_{\text{final}}}{T_{\text{initial}}} = \frac{600}{300} = 2.0 \).

Award [1] for the correct answer B.

The energy of the emitted photon is equal to the difference in energy between the two states: \( E = E_{\text{initial}} - E_{\text{final}} = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV} \).

Award [1] for the correct answer B.

The fundamental frequency of a pipe open at both ends is \( f_{\text{open}} = \frac{v}{2L} \). The third harmonic of a pipe closed at one end is \( f_{\text{closed}} = \frac{3v}{4L'} \). Equating these gives \( \frac{v}{2L} = \frac{3v}{4L'} \), which simplifies to \( \frac{1}{2L} = \frac{3}{4L'} \). Rearranging for the ratio gives \( \frac{L}{L'} = \frac{4}{6} = \frac{2}{3} \).

Award [1] for the correct answer A.

According to Faraday's law, the magnitude of the induced electromotive force (emf) is given by \( \mathcal{E} = A \frac{\Delta B}{\Delta t} \), where \( A = \pi r^2 \) is the area of the circular coil. Initially, \( \mathcal{E} = \pi r^2 \frac{\Delta B}{\Delta t} \). With the changes, the new area is \( A' = \pi (2r)^2 = 4\pi r^2 \) and the new rate of change is \( 2\frac{\Delta B}{\Delta t} \). The new emf is \( \mathcal{E}' = 4\pi r^2 \cdot 2\frac{\Delta B}{\Delta t} = 8\mathcal{E} \). The ratio is 8.

Award [1] for the correct answer C.

Initial momentum of the first block is \( mv \) (taking the initial direction as positive). After rebounding, its momentum is \( -\frac{1}{3}mv \). The change in momentum of the first block is \( \Delta p = -\frac{1}{3}mv - mv = -\frac{4}{3}mv \). By conservation of momentum or Newton's third law, the magnitude of the impulse delivered to the second block is equal to the magnitude of the change in momentum of the first block, which is \( \frac{4}{3}mv \).

Award 1 mark for the correct answer C.

The energy supplied to raise the temperature is \( Q_1 = P t_1 = m c (T_2 - T_1) \), which gives \( P = \frac{m c (T_2 - T_1)}{t_1} \). The energy supplied to melt the substance is \( Q_2 = P t_2 = m L \). Substituting \( P \) into the second equation gives \( \frac{m c (T_2 - T_1) t_2}{t_1} = m L \). Simplifying for \( L \) yields \( L = c (T_2 - T_1) \frac{t_2}{t_1} \).

Award 1 mark for the correct answer B.

The current in the circuit is \( I = \frac{E}{R+r} \), and the power dissipated in \( R \) is \( P_R = I^2 R = \frac{E^2 R}{(R+r)^2} \). When \( R = r \), \( P = \frac{E^2 r}{(2r)^2} = \frac{E^2}{4r} \). When \( R = 3r \), the new power is \( P' = \frac{E^2 (3r)}{(3r+r)^2} = \frac{3 E^2 r}{16 r^2} = \frac{3 E^2}{16 r} \). Comparing the two expressions, \( P' = \frac{3}{4} \left( \frac{E^2}{4r} \right) = \frac{3}{4} P \).

Award 1 mark for the correct answer C.

The average kinetic energy of gas molecules is directly proportional to the absolute temperature, so the average speed is proportional to \( \sqrt{T} \). Since the temperature increases by a factor of 1.44, the average speed increases by \( \sqrt{1.44} = 1.2 \). For a gas at a constant volume, pressure is directly proportional to temperature (\( P \propto T \)), so the pressure increases by a factor of 1.44.

Award 1 mark for the correct answer B.

The energy difference between levels is related to frequency by \( \Delta E = hf \) and to wavelength by \( \Delta E = \frac{hc}{\lambda} \). The total energy change for the direct transition from \( E_3 \) to \( E_1 \) is the sum of the individual transitions: \( \Delta E_{3 \to 1} = \Delta E_{3 \to 2} + \Delta E_{2 \to 1} = h f + \frac{hc}{\lambda} \). The frequency \( f_{3 \to 1} \) of the photon emitted in this direct transition is \( \frac{\Delta E_{3 \to 1}}{h} = f + \frac{c}{\lambda} \).

Award 1 mark for the correct answer B.

For single slit diffraction, the condition for the first minimum is \( b \sin\theta = \lambda \), which gives \( \sin\theta = \frac{\lambda}{b} \). If the slit width becomes \( b' = \frac{b}{2} \) and the wavelength becomes \( \lambda' = 2\lambda \), the sine of the new angle is \( \sin\theta' = \frac{\lambda'}{b'} = \frac{2\lambda}{b/2} = 4\frac{\lambda}{b} = 4\sin\theta \).

Award 1 mark for the correct answer C.

When unpolarized light of intensity \( I_0 \) passes through the first polarizer, it becomes vertically polarized with intensity \( I_1 = \frac{1}{2} I_0 \). According to Malus's Law, the intensity after passing through the second polarizer is \( I_2 = I_1 \cos^2\theta = \left(\frac{1}{2} I_0\right) \left(\frac{3}{5}\right)^2 = \frac{1}{2} I_0 \cdot \frac{9}{25} = \frac{9}{50} I_0 \).

Award 1 mark for the correct answer B.

The total binding energy of the initial nucleus \( \text{X} \) is \( 240 \times 7.6\text{ MeV} = 1824\text{ MeV} \). The combined binding energy of the two product nuclei \( \text{Y} \) is \( 2 \times (120 \times 8.5\text{ MeV}) = 240 \times 8.5\text{ MeV} = 2040\text{ MeV} \). The energy released is the difference in total binding energies between the products and the reactant: \( 2040\text{ MeV} - 1824\text{ MeV} = 216\text{ MeV} \).

Award 1 mark for the correct answer B.

The volume of water striking the wall per second is \(\frac{\Delta V}{\Delta t} = A v\). Thus, the mass of water striking the wall per second is \(\frac{\Delta m}{\Delta t} = \rho A v\). Since the water loses all of its horizontal velocity, the change in horizontal velocity is \(\Delta v = v\). According to Newton's second law, the force is equal to the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} \Delta v = (\rho A v) v = \rho A v^2\).

[1 mark] Award for correct choice B.

The luminosity of a star is given by the Stefan-Boltzmann law: \(L = \sigma A T^4 = \sigma (4\pi R^2) T^4\). Therefore, \(L \propto R^2 T^4\). The ratio of the luminosities is: \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = 3^2 \times 2^4 = 9 \times 16 = 144\).

[1 mark] Award for correct choice D.

The equivalent resistance of the two parallel resistors is \(R/2\). The total resistance of the circuit is \(R_{\text{total}} = R + R/2 = 1.5R = \frac{3}{2}R\). The total current in the circuit (which is the current through the third resistor) is \(I = \frac{V}{R_{\text{total}}} = \frac{2V}{3R}\). The power dissipated in this third resistor is \(P = I^2 R = \left(\frac{2V}{3R}\right)^2 R = \frac{4V^2}{9R^2} R = \frac{4V^2}{9R}\).

[1 mark] Award for correct choice A.

For a fixed mass of an ideal gas at constant volume, Gay-Lussac's law states that \(P \propto T\), where \(T\) is the absolute temperature in Kelvin. First, convert the temperatures to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Thus, the temperature is doubled. Since pressure is proportional to absolute temperature, the pressure also doubles: \(P_2 = 2P\).

[1 mark] Award for correct choice A.

For single slit diffraction, the first minimum on either side of the central maximum occurs at an angle \(\theta\) such that \(\sin\theta \approx \theta = \frac{\lambda}{b}\). The central maximum extends from the first minimum on one side to the first minimum on the other side. Therefore, the total angular width is \(2\theta = \frac{2\lambda}{b}\).

[1 mark] Award for correct choice B.

Emission occurs when an electron transitions from a higher energy level to a lower one, so transition 1 to 3 (which is absorption) is incorrect. The energy of the emitted photon is given by \(E = \frac{hc}{\lambda}\). Thus, the longest wavelength corresponds to the smallest energy difference. The energy differences for emission are: \(3 \rightarrow 2\) is \(3.0\text{ eV}\); \(3 \rightarrow 1\) is \(10.0\text{ eV}\); \(2 \rightarrow 1\) is \(7.0\text{ eV}\). The smallest energy transition is \(3.0\text{ eV}\) (from level 3 to level 2), which produces the longest wavelength.

[1 mark] Award for correct choice A.

(a) The Rayleigh criterion states that two sources are just resolved when the central maximum of the diffraction pattern of one source falls on the first minimum of the diffraction pattern of the other. (b) Using the formula for a circular aperture: \(\theta_{\text{min}} = 1.22 \frac{\lambda}{D}\). Substituting the values: \(\theta_{\text{min}} = 1.22 \times \frac{600 \times 10^{-9}\text{ m}}{0.50\text{ m}} = 1.464 \times 10^{-6}\text{ rad} \approx 1.5 \times 10^{-6}\text{ rad}\). (c) The physical separation \(s\) is related to the angular separation \(\theta\) and distance \(d\) by \(s = \theta \times d\). Substituting the values: \(s = 1.464 \times 10^{-6}\text{ rad} \times 4.2 \times 10^{16}\text{ m} = 6.1488 \times 10^{10}\text{ m} \approx 6.1 \times 10^{10}\text{ m}\). (d) First, find the grating spacing \(d\): \(d = \frac{10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m}\). Using the diffraction grating equation: \(d \sin\theta = n \lambda\). For the highest order, \(\sin\theta \le 1\). Therefore, \(n \le \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}\text{ m}}{600 \times 10^{-9}\text{ m}} = 3.33\). Since \(n\) must be an integer, the highest order observed is \(n = 3\).

(a) [1 mark] Reference to the diffraction patterns (central maximum / first minimum) of the two sources. [1 mark] Condition of overlap: central maximum of one falls on the first minimum of the other. (b) [1 mark] Correct substitution into the circular aperture formula. [1 mark] Correct final answer with correct units. (c) [1 mark] Correct substitution of \(s = \theta d\). [1 mark] Correct final answer with correct units. (d) [1 mark] Calculation of grating spacing \(d\). [1 mark] Use of the diffraction grating equation with \(\sin\theta = 1\) (or state \(\sin\theta \le 1\)). [1 mark] Calculation showing \(n = 3.33\). [1 mark] Correctly rounding down to the integer value \(n = 3\).

(a) Using the conservation of linear momentum: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v\). Substituting the given values: \(2.5 \times 4.0 + 1.5 \times 0 = (2.5 + 1.5) v \Rightarrow 10 = 4.0 v \Rightarrow v = 2.5\text{ m s}^{-1}\). (b) Initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (2.5)(4.0)^2 = 20\text{ J}\). Final kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (4.0)(2.5)^2 = 12.5\text{ J}\). Kinetic energy loss: \(\Delta E_k = E_{ki} - E_{kf} = 20 - 12.5 = 7.5\text{ J}\). (c) The impulse on the second block equals its change in momentum: \(I = \Delta p_2 = m_2 v - 0 = 1.5 \times 2.5 = 3.75\text{ N s}\). The average force is \(F_{\text{avg}} = \frac{\Delta p_2}{\Delta t} = \frac{3.75}{0.080} = 46.875\text{ N} \approx 47\text{ N}\). (d) Newton's third law states that the force exerted by block 1 on block 2 is equal in magnitude and opposite in direction to the force exerted by block 2 on block 1 (\(F_{12} = -F_{21}\)). According to Newton's second law, force is the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)). Since \(\Delta p_1 + \Delta p_2 = 0\) (as the internal forces cancel out) and there is no net external force, the total momentum remains constant.

(a) [1 mark] Application of momentum conservation equation. [1 mark] Correct calculation of the final velocity. (b) [1 mark] Calculation of initial kinetic energy. [1 mark] Calculation of final kinetic energy. [1 mark] Correct calculation of energy loss (7.5 J). (c) [1 mark] Calculation of change in momentum of the second block. [1 mark] Application of force-momentum relationship. [1 mark] Correct calculation of average force with unit. (d) [1 mark] Reference to Newton's third law (equal and opposite internal forces). [1 mark] Reference to Newton's second law (force as rate of change of momentum) or absence of external forces.

(a) Energy supplied \(Q_1 = P \times t_1\). Time \(t_1 = 3.0 \times 60 = 180\text{ s}\). So, \(Q_1 = 120\text{ W} \times 180\text{ s} = 21,600\text{ J} \approx 2.2 \times 10^4\text{ J}\). (b) Using \(Q_1 = m c \Delta T\), where \(\Delta T = 80 - 20 = 60\text{ K}\). Rearranging for specific heat capacity \(c\): \(c = \frac{Q_1}{m \Delta T} = \frac{21,600}{0.40 \times 60} = \frac{21,600}{24} = 900\text{ J kg}^{-1}\text{ K}^{-1}\). (c) Energy supplied during the phase change: \(Q_2 = P \times t_2\). Time \(t_2 = 5.0 \times 60 = 300\text{ s}\). So, \(Q_2 = 120\text{ W} \times 300\text{ s} = 36,000\text{ J}\). Using \(Q_2 = m L_f\), where \(L_f\) is the specific latent heat of fusion. Rearranging for \(L_f\): \(L_f = \frac{Q_2}{m} = \frac{36,000}{0.40} = 90,000\text{ J kg}^{-1} = 9.0 \times 10^4\text{ J kg}^{-1}\). (d) Some thermal energy is lost to the surroundings instead of being transferred directly to the substance. Consequently, the actual energy absorbed by the substance is less than the calculated energy supplied by the heater, resulting in an overestimation of the latent heat of fusion.

(a) [1 mark] Conversion of minutes to seconds. [1 mark] Correct calculation of energy (21,600 J or 22,000 J). (b) [1 mark] Calculation of temperature change (60 K or 60 °C). [1 mark] Correct substitution into specific heat equation. [1 mark] Correct final answer with correct units. (c) [1 mark] Calculation of energy supplied during melting (36,000 J). [1 mark] Substitution into latent heat equation. [1 mark] Correct final answer with correct units. (d) [1 mark] Stating that energy is lost to the surroundings / container. [1 mark] Explaining that this causes calculated energy to be higher than the energy actually used to melt the substance.

(a) The diagram must show a cell (with EMF \(\varepsilon\) and internal resistance \(r\) in series), a variable resistor, and an ammeter connected in a single loop (series). The voltmeter must be connected in parallel across the terminals of the cell. (b) The voltmeter measures the terminal potential difference \(V = 5.1\text{ V}\). The external circuit consists of the variable resistor \(R\) and the ammeter resistance \(R_A\) in series. Thus, \(V = I (R + R_A)\). Substituting values: \(5.1 = I (2.5 + 0.50) \Rightarrow 5.1 = 3.0 I \Rightarrow I = \frac{5.1}{3.0} = 1.7\text{ A}\). (c) The terminal potential difference is given by \(V = \varepsilon - I r\). Substituting values: \(5.1 = 6.0 - 1.7 r \Rightarrow 1.7 r = 0.9 \Rightarrow r = \frac{0.9}{1.7} \approx 0.529\ \Omega \approx 0.53\ \Omega\). (d) The new total resistance in the circuit is \(R_{\text{total}} = R + R_A + r = 5.5 + 0.50 + 0.529 = 6.529\ \Omega\). The new current is \(I_{\text{new}} = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.0}{6.529} \approx 0.919\text{ A}\). The new voltmeter reading (terminal potential difference) is \(V_{\text{new}} = \varepsilon - I_{\text{new}} r = 6.0 - (0.919 \times 0.529) = 6.0 - 0.486 = 5.514\text{ V} \approx 5.5\text{ V}\). Alternately: \(V_{\text{new}} = I_{\text{new}} (R + R_A) = 0.919 \times (5.5 + 0.50) = 5.514\text{ V}\).

(a) [1 mark] Series loop with cell, variable resistor, and ammeter using correct standard symbols. [1 mark] Voltmeter connected in parallel across the cell. (b) [1 mark] Identification of total external resistance \(R + R_A = 3.0\ \Omega\). [1 mark] Application of Ohm's law to show current is 1.7 A. (c) [1 mark] Use of formula \(V = \varepsilon - Ir\) or equivalent. [1 mark] Substitution of values into formula. [1 mark] Correct calculation of \(r\) as 0.53 ohms (accept 0.53 or 0.5). (d) [1 mark] Calculation of new total resistance (6.53 ohms). [1 mark] Calculation of new current (0.92 A). [1 mark] Correct final voltmeter reading (5.5 V).

(a) The ground state is the lowest possible energy state of an electron within an atom, representing the state where the electron is most tightly bound to the nucleus. (b) The energy difference for the transition from \(E_4\) to \(E_2\) is \(\Delta E = E_4 - E_2 = -2.0\text{ eV} - (-8.0\text{ eV}) = 6.0\text{ eV}\). Convert this energy to Joules: \(\Delta E = 6.0 \times 1.60 \times 10^{-19}\text{ J} = 9.6 \times 10^{-19}\text{ J}\). Using \(\Delta E = \frac{hc}{\lambda}\), we find: \(\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{9.6 \times 10^{-19}\text{ J}} = 2.07 \times 10^{-7}\text{ m} \approx 2.1 \times 10^{-7}\text{ m}\). (c) For a photon to be absorbed by an atom in its ground state (\(E_1 = -18.0\text{ eV}\)), its energy must exactly match the transition energy to a higher allowed state. Let us check the possible transitions: to state 2: \(\Delta E_{1\to2} = -8.0 - (-18.0) = 10.0\text{ eV}\); to state 3: \(\Delta E_{1\to3} = -4.5 - (-18.0) = 13.5\text{ eV}\); to state 4: \(\Delta E_{1\to4} = -2.0 - (-18.0) = 16.0\text{ eV}\). Since the photon energy of \(13.5\text{ eV}\) matches exactly the energy difference to transition the electron from \(E_1\) to \(E_3\), the photons can be absorbed. (d) Ionization requires removing the electron completely, which means transitioning it from its ground state \(E_1 = -18.0\text{ eV}\) to the free state of energy \(0\text{ eV}\). The minimum energy required is \(E = 18.0\text{ eV} = 18.0 \times 1.60 \times 10^{-19}\text{ J} = 2.88 \times 10^{-18}\text{ J}\). Using \(E = hf\), we find: \(f = \frac{E}{h} = \frac{2.88 \times 10^{-18}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 4.34 \times 10^{15}\text{ Hz} \approx 4.3 \times 10^{15}\text{ Hz}\).

(a) [1 mark] Identification of ground state as the lowest energy level of an electron in an atom. (b) [1 mark] Calculation of transition energy (6.0 eV). [1 mark] Conversion of energy into Joules. [1 mark] Correct calculation of wavelength with unit. (c) [1 mark] Stating that photon absorption requires an exact match with a transition energy. [1 mark] Calculation of transition energies from the ground state (showing \(E_3 - E_1 = 13.5\text{ eV}\)). [1 mark] Concluding that absorption is possible because of this exact match. (d) [1 mark] Stating the ionization energy is 18.0 eV. [1 mark] Converting ionization energy to Joules. [1 mark] Correct calculation of frequency with unit.

Part (a):
The circuit should show the cell connected in series with an ammeter and a variable resistor. A voltmeter must be connected in parallel across the cell (or across the variable resistor).

Part (b):
The relationship is given by \(V = \mathcal{E} - Ir\). comparing this to the equation of a straight line, \(y = mx + c\):
- The vertical intercept (when \(I = 0\)) is equal to the emf \(\mathcal{E}\).
- The gradient (slope) of the line is equal to \(-r\), so the internal resistance is the magnitude of the gradient.

Part (c)(i):
Power \(P = V \times I = 1.12 \times 0.50 = 0.56\text{ W}\).
Fractional uncertainty in current: \(\frac{\Delta I}{I} = \frac{0.02}{0.50} = 0.040\) (or \(4.0\%\)).
Fractional uncertainty in voltage: \(\frac{\Delta V}{V} = \frac{0.01}{1.12} \approx 0.0089\) (or \(0.89\%\)).
Fractional uncertainty in power: \(\frac{\Delta P}{P} = \frac{\Delta I}{I} + \frac{\Delta V}{V} = 0.040 + 0.0089 = 0.0489\) (or \(4.89\%\)).
Absolute uncertainty in power: \(\Delta P = 0.56 \times 0.0489 \approx 0.0274\text{ W}\).
Rounding the uncertainty to 1 significant figure gives \(\Delta P = 0.03\text{ W}\).
Therefore, \(P = 0.56 \pm 0.03\text{ W}\).

Part (c)(ii):
Since \(R = \frac{V}{I}\), the percentage uncertainty in \(R\) is the sum of the percentage uncertainties of \(V\) and \(I\):
\(\%\Delta R = \%\Delta V + \%\Delta I = 0.89\% + 4.0\% \approx 4.9\%\) (or \(5\%\)).

Part (a):
- Voltmeter connected in parallel across the cell or variable resistor AND ammeter in series. [1 mark]
- Variable resistor included in series to allow variation of current. [1 mark]

Part (b):
- Identifies \(\mathcal{E}\) as the y-intercept. [1 mark]
- Identifies \(r\) as the negative gradient / magnitude of the gradient. [1 mark]

Part (c)(i):
- Calculates \(P = 0.56\text{ W}\). [0.5 marks]
- Sums the fractional/percentage uncertainties: \(0.040 + 0.0089 = 0.0489\) (or \(4.9\%\)). [1 mark]
- Obtains absolute uncertainty \(\pm 0.03\text{ W}\) (accept \(\pm 0.027\text{ W}\)) and states final answer with correct units: \(0.56 \pm 0.03\text{ W}\). [1 mark]

Part (c)(ii):
- Correctly states \(4.9\%\) or \(5\%\). [1 mark]

Part (a):
The ideal gas law is \(pV = nRT\).
The relationship between absolute temperature \(T\) (in K) and temperature \(\theta\) (in \(^\circ\text{C}\)) is \(T = \theta + T_0\), where \(T_0 = 273.15\text{ K}\).
Substituting this into the gas law:
\(V = \frac{nR}{p}(\theta + T_0) = \left(\frac{nR}{p}\right)\theta + \frac{nRT_0}{p}\).
Comparing this to \(V = m\theta + c\), we get \(m = \frac{nR}{p}\) and \(c = \frac{nRT_0}{p}\).
The ratio \(-\frac{c}{m} = -\left(\frac{nRT_0/p}{nR/p}\right) = -T_0\), which physically represents absolute zero in degrees Celsius.

Part (b):
At absolute zero, the volume \(V\) of an ideal gas is theoretically zero.
Using the line of best fit:
\(0 = 0.150\,\theta + 41.0\)
\(\theta = -\frac{41.0}{0.150} \approx -273.33^\circ\text{C}\).
Therefore, absolute zero is estimated to be \(-273^\circ\text{C}\) (or \(-273.3^\circ\text{C}\)).

Part (c):
A possible systematic error is the end correction of the capillary tube (e.g., the bottom of the tube is hemispherical or not perfectly flat, or the liquid plug has a curved meniscus).
This results in an overestimate of the volume by a constant value \(\Delta V\).
Since \(V_{measured} = V_{actual} + \Delta V\), the constant \(c\) (the y-intercept) will be larger than its true value.
Since absolute zero is calculated as \(-\frac{c}{m}\), a larger \(c\) leads to a more negative value for absolute zero, explaining why the experimental value differs from the accepted value.

Part (a):
- Substitutes \(T = \theta + T_0\) (or \(\theta + 273\)) into the ideal gas law \(pV = nRT\). [1 mark]
- Rearranges equation to the form \(V = \left(\frac{nR}{p}\right)\theta + \frac{nRT_0}{p}\). [1 mark]
- Identifies \(m = \frac{nR}{p}\) and \(c = \frac{nRT_0}{p}\). [0.5 marks]
- States that \(-\frac{c}{m}\) represents absolute zero in degrees Celsius (or \(-T_0\)). [1 mark]

Part (b):
- Equates \(V\) to \(0\) (or states \(\theta = -\frac{c}{m}\)). [1 mark]
- Obtains \(-273^\circ\text{C}\) or \(-273.3^\circ\text{C}\) (negative sign and Celsius unit must be present). [1 mark]

Part (c):
- Identifies a valid systematic error (e.g., end-correction of the tube, gas leakage, thermal expansion of the tube itself). [1 mark]
- Explains its effect on the intercept \(c\) (or gradient \(m\)) and how this shifts the calculated ratio \(-\frac{c}{m}\) away from the true value. [1 mark]

(a) A standard candle is an astronomical object with a known luminosity.

(b) 1. The period of variation of the Cepheid's luminosity is measured from its light curve.
2. The period-luminosity relationship (a known empirical relation for Cepheid variables) is used to find the absolute luminosity \(L\) of the star.
3. By measuring the apparent brightness \(b\) from Earth, the distance \(d\) can be calculated using the relation \(b = \frac{L}{4\pi d^2}\).

(c)(i) Using the relation \(b = \frac{L}{4\pi d^2}\):
\(d = \sqrt{\frac{L}{4\pi b}}\)
\(d = \sqrt{\frac{3.1 \times 10^{30}}{4\pi \times 1.2 \times 10^{-11}}} = \sqrt{\frac{3.1 \times 10^{30}}{1.508 \times 10^{-10}}} \approx 1.43 \times 10^{20} \text{ m}\)

Convert meters to parsecs:
\(1 \text{ pc} = 3.09 \times 10^{16} \text{ m}\)
\(d = \frac{1.43 \times 10^{20}}{3.09 \times 10^{16}} \approx 4.64 \times 10^3 \text{ pc}\) (or \(4.6 \times 10^3 \text{ pc}\)).

(c)(ii) 1. Beyond a certain distance, Cepheid variables become too faint to be resolved individually from other stars in distant galaxies.
2. Interstellar dust or gas can absorb and scatter light (interstellar extinction), making the star appear dimmer than it is, leading to an overestimation of distance.

(a) Award [1] for: An astronomical object of known luminosity (or absolute magnitude).

(b) Award [1] for stating that the period of variation is measured. Award [1] for stating that the period determines the luminosity using the period-luminosity relationship. Award [1] for stating that distance is calculated using the relationship between luminosity and apparent brightness.

(c)(i) Award [1] for substitution of values into the inverse square law equation to find distance in meters: \(1.43 \times 10^{20} \text{ m}\). Award [1] for dividing by \(3.09 \times 10^{16}\) to convert to parsecs. Award [1] for correct final answer: \(4.6 \times 10^3 \text{ pc}\) (accept range \(4.6 \times 10^3 \text{ pc}\) to \(4.64 \times 10^3 \text{ pc}\)).

(c)(ii) Award [1] for identifying a valid limitation (e.g., interstellar dust, faintness at great distances, resolution limit). Award [2] max for a detailed explanation of how this affects the measurement: If faintness: At extremely large distances, individual Cepheids cannot be distinguished from background stars, making their periods impossible to measure. If interstellar dust: Dust absorbs/scatters light, making the star appear dimmer, which causes the calculated distance to be overestimated.

(a) 1. Cosmic Microwave Background (CMB) radiation.
2. Redshift of light from distant galaxies (expansion of space).

(b) 1. In the hot, dense early universe, radiation was in thermal equilibrium with matter, following a blackbody spectrum at a very high temperature.
2. As the universe expanded, space stretched, which also stretched the wavelength of this relic radiation.
3. This redshifted the radiation into the microwave region of the spectrum, corresponding to a blackbody temperature of about \(2.7 \text{ K}\) today, which is highly isotropic.

(c)(i) Redshift \(z\) is given by:
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{\lambda - \lambda_0}{\lambda_0}\)
\(z = \frac{702.1 - 656.3}{656.3} = \frac{45.8}{656.3} \approx 0.0698\) (or \(0.070\))

(c)(ii) Using \(v = z c\):
\(v = 0.0698 \times 3.00 \times 10^8 \text{ m s}^{-1} \approx 2.09 \times 10^7 \text{ m s}^{-1}\) (or \(2.1 \times 10^4 \text{ km s}^{-1}\))

(c)(iii) Using Hubble's Law: \(v = H_0 d \implies d = \frac{v}{H_0}\)
Converting \(v\) to \(\text{km s}^{-1}\):
\(d = \frac{2.09 \times 10^4}{70} \approx 299 \text{ Mpc}\) (or \(300 \text{ Mpc}\)).

(a) Award [1] each for any two of the following: Cosmic Microwave Background (CMB) radiation; Cosmological redshift / expansion of space; Relative abundance of light elements (He, D, Li).

(b) Award [1] for stating that the early universe was very hot, dense, and filled with short-wavelength radiation. Award [1] for stating that the expansion of space stretched the wavelengths of these photons. Award [1] for connecting the stretched wavelengths to the observed microwave region / low temperature of \(2.7 \text{ K}\) isotropic radiation today.

(c)(i) Award [1] for calculating \(\Delta \lambda = 45.8 \text{ nm}\) or showing substitution. Award [1] for correct calculation of \(z = 0.070\) (or \(0.0698\)).

(c)(ii) Award [1] for correct calculation of velocity: \(2.1 \times 10^7 \text{ m s}^{-1}\) (or \(2.1 \times 10^4 \text{ km s}^{-1}\)). Allow ECF from (c)(i).

(c)(iii) Award [1] for correct substitution into \(d = \frac{v}{H_0}\) with consistent units. Award [1] for correct distance calculation: \(300 \text{ Mpc}\) (or \(299 \text{ Mpc}\)). Allow ECF from (c)(ii).