2025 Edexcel IAL Physics (YPH11) 模擬試題連答案詳解

The radius of a circular path for a charged particle in a magnetic field is given by \(r = \frac{mv}{Bq}\).
Since the momentum \(p = mv\) and kinetic energy \(E_k = \frac{p^2}{2m}\), we can write the momentum as \(p = \sqrt{2mE_k}\).
Substituting this into the radius equation gives:
\(r = \frac{\sqrt{2mE_k}}{Bq}\).
Since both particles have the same kinetic energy \(E_k\), the same charge magnitude \(e\), and are in the same magnetic field \(B\), the radius is directly proportional to the square root of the mass:
\(r \propto \sqrt{m}\).
Therefore, the ratio of the radius of the proton's path to the electron's path is \(\frac{r_p}{r_e} = \sqrt{\frac{m_p}{m_e}}\).

1 mark for the correct option B.

According to Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\), so temperature \(T \propto \frac{1}{\lambda_{\text{max}}}\).
Thus, \(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = 2\).
According to Stefan-Boltzmann law, luminosity \(L = 4\pi r^2 \sigma T^4\).
Since both stars have the same luminosity \(L_X = L_Y\):
\(r_X^2 T_X^4 = r_Y^2 T_Y^4 \implies \left(\frac{r_X}{r_Y}\right)^2 = \left(\frac{T_Y}{T_X}\right)^4\).
Substituting \(\frac{T_Y}{T_X} = \frac{1}{2}\):
\left(\frac{r_X}{r_Y}\right)^2 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \implies \frac{r_X}{r_Y} = \sqrt{\frac{1}{16}} = \frac{1}{4}\).

1 mark for the correct option B.

The total energy in simple harmonic motion is given by \(E_T = \frac{1}{2} k A^2\).
The potential energy is \(E_p = \frac{1}{2} k x^2\).
The kinetic energy is \(E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2)\).
We are given that \(E_k = 3 E_p\):
\(\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)\)
\(A^2 - x^2 = 3x^2\)
\(A^2 = 4x^2\)
\(x^2 = \frac{A^2}{4} \implies x = \pm \frac{A}{2}\).

1 mark for the correct option C.

The pion-minus is a meson consisting of a down quark and an up antiquark (\(d\bar{u}\)). It decays into leptons, which do not experience the strong interaction. This transformation must involve the weak interaction, as quark flavour is changed and leptons are produced.

1 mark for the correct option C.

The absolute temperature \(T\) of an ideal gas is directly proportional to the average kinetic energy of its molecules: \(E_k = \frac{3}{2} k T = \frac{1}{2} m \langle c^2 \rangle\), where \(\langle c^2 \rangle\) is the mean square speed. Since \(T\) is doubled, the mean square speed of the molecules is doubled. The root-mean-square speed \(v_{\text{rms}} = \sqrt{\langle c^2 \rangle}\) increases by a factor of \(\sqrt{2}\), not \(2\).

1 mark for the correct option A.

After three half-lives, the number of remaining nuclei is:
\(N = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\).
The number of nuclei that have decayed is:
\(N_d = N_0 - N = N_0 - \frac{N_0}{8} = \frac{7N_0}{8}\).
The ratio of decayed to remaining nuclei is:
\(\frac{N_d}{N} = \frac{7N_0/8}{N_0/8} = 7\).

1 mark for the correct option A.

At the highest point of the vertical circle, both the tension \(T\) and the weight \(mg\) act downwards towards the centre of the circle, providing the centripetal force:
\(F_{\text{net}} = T + mg = \frac{mv^2}{L}\).
Since \(T = mg\):
\(mg + mg = \frac{mv^2}{L} \implies 2mg = \frac{mv^2}{L} \implies v^2 = 2gL \implies v = \sqrt{2gL}\).

1 mark for the correct option B.

The initial charge on the capacitor is \(Q = CV\).
When connected in parallel with an uncharged capacitor of capacitance \(2C\), the equivalent capacitance of the system becomes \(C_{\text{total}} = C + 2C = 3C\).
Since the system is isolated, the total charge \(Q\) remains conserved.
The final total electrostatic energy stored is:
\(E_{\text{final}} = \frac{Q^2}{2C_{\text{total}}} = \frac{(CV)^2}{2(3C)} = \frac{C^2 V^2}{6C} = \frac{1}{6} C V^2\).

1 mark for the correct option C.

Using the relationship between kinetic energy and accelerating voltage: \(E_k = qV\). Since \(E_k = \frac{p^2}{2m}\), we can write the momentum as \(p = \sqrt{2mqV}\). In a magnetic field, the radius of the path is given by \(r = \frac{p}{qB} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}\). For a constant potential difference \(V\) and magnetic flux density \(B\), the radius is proportional to \(\sqrt{\frac{m}{q}}\). Therefore, the ratio of the radii is \(\frac{r_{\alpha}}{r_p} = \sqrt{\frac{m_{\alpha}}{m_p} \times \frac{q_p}{q_{\alpha}}} = \sqrt{4 \times \frac{1}{2}} = \sqrt{2}\).

1 mark for the correct option C.

The original pendulum period is \(T = 2\pi\sqrt{\frac{L}{g}}\). For the modified pendulum, one half of the oscillation is a standard swing of length \(L\), which takes half of the original period: \(t_1 = \frac{1}{2} T\). The other half of the oscillation has an effective length of \(L' = 0.25L\). A full pendulum of this length would have a period of \(T' = 2\pi\sqrt{\frac{0.25L}{g}} = 0.5 T\). Thus, the half-cycle for this shorter swing takes \(t_2 = \frac{1}{2} T' = 0.25 T\). The total period of the modified pendulum is \(t_1 + t_2 = 0.5 T + 0.25 T = 0.75 T\).

1 mark for the correct option C.

(a) The electrical work done on the ion is equal to its gain in kinetic energy:
\(qV = \frac{1}{2}mv^2\)
Rearranging this formula for \(v\):
\(v^2 = \frac{2qV}{m} \implies v = \sqrt{\frac{2qV}{m}}\)

(b) Substituting the values into the equation:
\(v = \sqrt{\frac{2 \times (1.60 \times 10^{-19}\text{ C}) \times 2500\text{ V}}{3.98 \times 10^{-26}\text{ kg}}}\)
\(v = \sqrt{2.01 \times 10^{10}} = 1.42 \times 10^5\text{ m s}^{-1}\)

(c) The magnetic force acts as the centripetal force:
\(Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}\)
Substituting the values:
\(r = \frac{3.98 \times 10^{-26}\text{ kg} \times 1.42 \times 10^5\text{ m s}^{-1}}{0.18\text{ T} \times 1.60 \times 10^{-19}\text{ C}}\)
\(r = 0.196\text{ m} \approx 0.20\text{ m}\)

(d) Magnesium-26 has a greater mass than magnesium-24 but the same charge. Since \(r = \frac{1}{B}\sqrt{\frac{2mV}{q}}\), the radius of the circular path is directly proportional to \(\sqrt{m}\). Therefore, the \(^{26}\text{Mg}^+\) ions will travel in a circular path with a larger radius (less curvature).

**Part (a) [2 Marks]**
- Equates electrical potential energy to kinetic energy: \(qV = \frac{1}{2}mv^2\) (1)
- Rearranges successfully to show the given equation (1)

**Part (b) [2 Marks]**
- Correct substitution of values into the formula (1)
- Correct final speed: \(1.42 \times 10^5\text{ m s}^{-1}\) (accept \(1.4 \times 10^5\text{ m s}^{-1}\)) (1)

**Part (c) [3 Marks]**
- Identifies magnetic force equals centripetal force: \(Bqv = \frac{mv^2}{r}\) (1)
- Rearranges to \(r = \frac{mv}{Bq}\) and substitutes values (1)
- Correct radius: \(0.20\text{ m}\) (accept \(0.196\text{ m}\) to \(0.20\text{ m}\)) (1)

**Part (d) [2 Marks]**
- Explains that the greater mass of \(^{26}\text{Mg}^+\) means it has a larger radius / less deflection (1)
- Explains this using the relationship \(r \propto \sqrt{m}\) or by noting momentum is greater (1)

(a) Using Wien's Law:
\(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\)
\(T = \frac{2.90 \times 10^{-3}\text{ m K}}{300 \times 10^{-9}\text{ m}} = 9.67 \times 10^3\text{ K}\) (or \(9670\text{ K}\))

(b) Using Stefan-Boltzmann's Law:
\(L = 4\pi R^2 \sigma T^4\)
Rearranging for \(R\):
\(R = \sqrt{\frac{L}{4\pi\sigma T^4}}\)
\(R = \sqrt{\frac{2.2 \times 10^{28}}{4\pi \times (5.67 \times 10^{-8}) \times (9670)^4}}\)
\(R = \sqrt{\frac{2.2 \times 10^{28}}{6.22 \times 10^9}} = \sqrt{3.54 \times 10^{18}} \approx 1.88 \times 10^9\text{ m}\), which is approximately \(1.9 \times 10^9\text{ m}\).

(c) First, convert the distance to metres:
\(d = 7.7 \times 3.09 \times 10^{16}\text{ m} = 2.38 \times 10^{17}\text{ m}\)
Then calculate radiation flux:
\(F = \frac{L}{4\pi d^2} = \frac{2.2 \times 10^{28}}{4\pi \times (2.38 \times 10^{17})^2} = 3.1 \times 10^{-8}\text{ W m}^{-2}\)

(d) The wavelengths of absorption lines in Vega's spectrum are compared with standard laboratory wavelengths of the same elements. If the lines are shifted to longer wavelengths (redshift), Vega is moving away from Earth. If they are shifted to shorter wavelengths (blueshift), Vega is moving towards Earth. The size of the shift determines the relative speed.

**Part (a) [2 Marks]**
- Recalls and uses \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\) (1)
- Correctly calculates \(T = 9.67 \times 10^3\text{ K}\) (1)

**Part (b) [3 Marks]**
- Recalls and uses \(L = 4\pi R^2 \sigma T^4\) (1)
- Correct substitution of values including calculated \(T\) (1)
- Obtains value of approximately \(1.9 \times 10^9\text{ m}\) (1)

**Part (c) [2 Marks]**
- Calculates distance in metres: \(2.38 \times 10^{17}\text{ m}\) (1)
- Correct radiation flux: \(3.1 \times 10^{-8}\text{ W m}^{-2}\) (1)

**Part (d) [2 Marks]**
- Explains comparing absorption lines with laboratory references to identify redshift/blueshift (1)
- Correlates redshift/blueshift with direction of motion (away/towards) (1)

(a) The equation is:
\(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + \beta^- + \bar{\nu}_e\)
(Accept \(^{0}_{-1}\text{e}\) for \(\beta^-\)

(b) The initial activity \(A_0\) of \(45\text{ g}\) of living carbon is:
\(A_0 = 15.0\text{ Bq/g} \times 45\text{ g} = 675\text{ Bq}\)
Calculate decay constant \(\lambda\):
\(\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{5730\text{ years}} = 1.21 \times 10^{-4}\text{ yr}^{-1}\)
Use the radioactive decay law:
\(A = A_0 e^{-\lambda t}\)
\(320 = 675 e^{-(1.21 \times 10^{-4})t}\)
\(\frac{320}{675} = e^{-(1.21 \times 10^{-4})t} \implies 0.474 = e^{-(1.21 \times 10^{-4})t}\)
\(\ln(0.474) = -(1.21 \times 10^{-4})t\)
\(-0.746 = -1.21 \times 10^{-4} t\)
\(t = \frac{0.746}{1.21 \times 10^{-4}} = 6165\text{ years} \approx 6.2 \times 10^3\text{ years}\)

(c) Assumptions/limitations:
1. It is assumed that the atmospheric ratio of Carbon-14 to Carbon-12 has remained constant over the last several thousand years, which may not be completely true due to solar activity or industrial combustion of fossil fuels.
2. The activity must be high enough to measure accurately; very old samples (\(> 50,000\) years) have too little Carbon-14 left, making detection highly inaccurate.

**Part (a) [2 Marks]**
- Balanced nucleon and proton numbers: Nitrogen-14 shown correctly (1)
- Identification of beta-minus particle and electron antineutrino (1)

**Part (b) [4 Marks]**
- Calculates initial activity \(A_0 = 675\text{ Bq}\) (1)
- Calculates decay constant \(\lambda = 1.21 \times 10^{-4}\text{ yr}^{-1}\) (1)
- Applies radioactive decay equation correctly (1)
- Obtains final age: \(6.2 \times 10^3\text{ years}\) (accept \(6100 - 6200\text{ years}\)) (1)

**Part (c) [3 Marks]**
- Identifies first limitation/assumption (e.g. constant ratio of C-14/C-12 in atmosphere) and explains it (1)
- Identifies second limitation (e.g. background radiation interference or age limit of around 50,000 years) and explains it (1)
- Clear physics terminology used throughout (1)

(a) At the top of the loop, both forces act vertically downwards:
- Weight of the cart (\(mg\)) acting downwards from the centre of mass.
- Normal contact force (\(R\)) from the track acting downwards.

(b) The centripetal force is provided by the sum of weight and normal contact force:
\(F_c = mg + R = \frac{mv^2}{r}\)
At minimum speed, the cart is on the verge of losing contact, so \(R = 0\).
\(mg = \frac{mv_{\text{min}}^2}{r} \implies v_{\text{min}} = \sqrt{gr}\)
\(v_{\text{min}} = \sqrt{9.81\text{ m s}^{-2} \times 12\text{ m}} = 10.8\text{ m s}^{-1} \approx 11\text{ m s}^{-1}\)

(c) At the bottom of the loop, the normal force \(R\) acts upwards and the weight \(mg\) acts downwards. The centripetal force is the net upward force:
\(R - mg = \frac{mv^2}{r}\)
\(R = mg + \frac{mv^2}{r}\)
\(R = (250 \times 9.81) + \frac{250 \times 24^2}{12}\)
\(R = 2452.5 + \frac{250 \times 576}{12} = 2452.5 + 12000 = 14452.5\text{ N} \approx 1.4 \times 10^4\text{ N}\)

**Part (a) [2 Marks]**
- Downward arrow representing Weight/gravity force correctly labelled (1)
- Downward arrow representing Normal contact force/reaction force from track correctly labelled (1)

**Part (b) [3 Marks]**
- Sets \(R = 0\) for minimum speed condition and writes \(mg = \frac{mv^2}{r}\) (1)
- Correct substitution of values (1)
- Correct final minimum speed: \(11\text{ m s}^{-1}\) (accept \(10.8\text{ m s}^{-1}\)) (1)

**Part (c) [4 Marks]**
- Identifies centripetal force equation at bottom: \(R - mg = \frac{mv^2}{r}\) (1)
- Rearranges to solve for \(R\) (1)
- Substitutes values correctly (1)
- Correct final normal force: \(1.4 \times 10^4\text{ N}\) (accept \(14\text{ kN}\) or \(14.5\text{ kN}\)) (1)

(a) Using the equation for frequency of a mass-spring system:
\(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
\(f = \frac{1}{2\pi}\sqrt{\frac{18}{0.45}} = \frac{1}{2\pi}\sqrt{40} = 1.006\text{ Hz} \approx 1.0\text{ Hz}\)

(b) Since the block is released from maximum displacement \(A = 0.12\text{ m}\) at \(t = 0\), the displacement can be modeled using a cosine function:
\(x = A \cos(\omega t)\)
\(\omega = 2\pi f = \sqrt{\frac{k}{m}} = \sqrt{40} \approx 6.32\text{ rad s}^{-1}\)
\(x = 0.12 \cos(6.3 t)\) (where \(x\) is in metres, \(t\) is in seconds).

(c) The maximum velocity is given by:
\(v_{\text{max}} = \omega A\)
\(v_{\text{max}} = 6.32\text{ rad s}^{-1} \times 0.12\text{ m} = 0.758\text{ m s}^{-1} \approx 0.76\text{ m s}^{-1}\)

(d) Total energy is constant throughout the oscillation.
- Potential energy is maximum at the extreme positions (\(x = \pm A\)), where displacement is maximum, and is zero at the equilibrium position (\(x = 0\)).
- Kinetic energy is maximum at the equilibrium position (\(x = 0\)), where the speed is maximum, and is zero at the extremes (\(x = \pm A\)).
As the block oscillates, energy is continuously transferred between potential and kinetic energy forms, with \(E_k + E_p = \text{constant}\).

**Part (a) [2 Marks]**
- Recalls and uses \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\) (1)
- Obtains a value of \(1.01\text{ Hz}\) and shows it rounds to \(1.0\text{ Hz}\) (1)

**Part (b) [2 Marks]**
- Uses cosine function due to release from maximum displacement (1)
- Correct values for amplitude \(0.12\) and angular frequency \(6.3\text{ rad s}^{-1}\) (1)

**Part (c) [2 Marks]**
- Recalls and uses \(v_{\text{max}} = \omega A\) (1)
- Correct calculation of \(0.76\text{ m s}^{-1}\) (1)

**Part (d) [3 Marks]**
- Explains that potential energy is maximum at displacement extremes (\(x = \pm 0.12\text{ m}\)) and zero at center (1)
- Explains that kinetic energy is maximum at equilibrium (\(x = 0\)) and zero at extremes (1)
- States that total mechanical energy remains constant (1)

(a) The thermal energy \(Q\) required to raise the temperature of the water:
\(Q = mc\Delta T = 1.2\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1} \times (100 - 20)\text{ K}\)
\(Q = 1.2 \times 4200 \times 80 = 4.032 \times 10^5\text{ J}\)
Since \(P = \frac{Q}{t}\):
\(t = \frac{Q}{P} = \frac{4.032 \times 10^5\text{ J}}{2200\text{ W}} = 183\text{ s} \approx 180\text{ s}\) (or \(3.1\) minutes).

(b) The electrical energy supplied during boiling:
\(E = P \times t = 2200\text{ W} \times 60\text{ s} = 1.32 \times 10^5\text{ J}\)
Mass vaporised \(m = 50\text{ g} = 0.050\text{ kg}\).
Using \(E = m L_v\):
\(L_v = \frac{E}{m} = \frac{1.32 \times 10^5\text{ J}}{0.050\text{ kg}} = 2.64 \times 10^6\text{ J kg}^{-1} \approx 2.6 \times 10^6\text{ J kg}^{-1}\).

(c) During fusion (melting), intermolecular bonds are only weakened or partially broken; the spacing between molecules changes very little, meaning minimal work is done against attractive forces. During vaporisation (boiling), intermolecular bonds must be completely broken and the spacing between molecules increases enormously, requiring a vast amount of work to overcome the attractive forces and expand against atmospheric pressure. Therefore, vaporisation requires much more energy per unit mass.

**Part (a) [3 Marks]**
- Uses \(Q = mc\Delta T\) with \(\Delta T = 80\text{ K}\) (1)
- Uses \(P = \frac{Q}{t}\) to find \(t\) (1)
- Correct final time: \(183\text{ s}\) or \(180\text{ s}\) (1)

**Part (b) [3 Marks]**
- Calculates electrical energy supplied: \(1.32 \times 10^5\text{ J}\) (1)
- Converts mass to kg: \(0.050\text{ kg}\) (1)
- Correct final latent heat: \(2.6 \times 10^6\text{ J kg}^{-1}\) (1)

**Part (c) [3 Marks]**
- States that melting only weakens bonds while boiling breaks them completely (1)
- Explains that the increase in molecular separation is much larger in vaporisation (1)
- Mentions that significant work must be done against intermolecular forces and atmospheric pressure during boiling (1)

(a) A typical production reaction is:
\(p + p \rightarrow p + p + \pi^0\)
Both incoming protons are conserved, and the kinetic energy of the collision is converted into the mass of the \(\pi^0\) meson.

(b) Charge calculation:
- Charge of \(u = +\frac{2}{3}e\)
- Charge of \(\bar{u} = -\frac{2}{3}e\) (opposite to the quark)
- Total charge = \(+\frac{2}{3}e - \frac{2}{3}e = 0\).

Baryon number calculation:
- Baryon number of any quark = \(+\frac{1}{3}\)
- Baryon number of any antiquark = \(-\frac{1}{3}\)
- Total baryon number \(B = +\frac{1}{3} - \frac{1}{3} = 0\).
These match the properties of the neutral pion.

(c) In a cyclotron:
- The uniform magnetic field is directed perpendicularly to the dees, exerting a force perpendicular to the velocity of the protons. This provides the centripetal force (\(Bqv = \frac{mv^2}{r}\)), causing the protons to move in circular paths within the dees.
- The alternating potential difference is applied across the gap between the dees. It accelerates the protons as they cross the gap. The frequency of the AC matches the circular orbit frequency (\(f = \frac{Bq}{2\pi m}\)), which is constant, ensuring the field reverses direction just as the protons reach the gap.

**Part (a) [2 Marks]**
- Left-hand side correct: \(p + p\) (1)
- Right-hand side showing conservation of charge and baryon number: \(p + p + \pi^0\) (1)

**Part (b) [3 Marks]**
- Correct charge calculation showing net charge is \(0\) (1)
- Correct baryon number calculation showing \(B = 0\) (1)
- Concludes clearly that this matches pion properties (1)

**Part (c) [4 Marks]**
- Explains magnetic field provides centripetal force perpendicular to motion (1)
- Explains electric field/potential difference accelerates protons across the gap (1)
- Explains that the frequency of the AC must be constant because orbital period is independent of speed (1)
- Mentions how the spiral path increases radius as velocity increases (1)

(a) Observation: Concentric bright and dark circles/rings are seen on a fluorescent screen placed behind the graphite film. This interference pattern can only be explained by wave behaviour, proving that moving electrons behave as waves.

(b) First, find the kinetic energy \(E_k\) gained by the electron:
\(E_k = eV = (1.60 \times 10^{-19}\text{ C}) \times 4500\text{ V} = 7.20 \times 10^{-16}\text{ J}\)
Next, calculate the momentum \(p\) of the electron:
\(E_k = \frac{p^2}{2m_e} \implies p = \sqrt{2 m_e E_k}\)
\(p = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (7.20 \times 10^{-16}\text{ J})} = \sqrt{1.312 \times 10^{-45}} = 3.62 \times 10^{-23}\text{ kg m s}^{-1}\)
Now, find the de Broglie wavelength:
\(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{3.62 \times 10^{-23}\text{ kg m s}^{-1}} = 1.83 \times 10^{-11}\text{ m} \approx 1.8 \times 10^{-11}\text{ m}\).

(c) The resolution of any microscope is limited by the diffraction of the probe used. Waves diffract (spread out) when passing through apertures or around obstacles of similar size to their wavelength, obscuring fine detail. The de Broglie wavelength of accelerated electrons (\(\sim 10^{-11}\text{ m}\)) is several orders of magnitude smaller than the wavelength of visible light (\(\sim 400 - 700\text{ nm}\)). Therefore, electrons experience significantly less diffraction, allowing much smaller details to be resolved.

**Part (a) [2 Marks]**
- Describes observing concentric rings/diffraction pattern on the screen (1)
- Concludes this shows wave-like behavior of electrons (1)

**Part (b) [4 Marks]**
- Calculates kinetic energy: \(7.20 \times 10^{-16}\text{ J}\) (1)
- Uses \(p = \sqrt{2mE_k}\) to find momentum: \(3.62 \times 10^{-23}\text{ kg m s}^{-1}\) (1)
- Uses \(\lambda = \frac{h}{p}\) with correct substitution (1)
- Correct final wavelength: \(1.8 \times 10^{-11}\text{ m}\) (accept \(1.83 \times 10^{-11}\text{ m}\)) (1)

**Part (c) [3 Marks]**
- Explains that resolution is limited by diffraction / diffraction obscures detail (1)
- Compares wavelengths: electron wavelength is much smaller than light wavelength (1)
- Explains that smaller wavelength results in less diffraction, thus higher resolution (1)

**(a)**
An electron of charge \(e\) accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[eV = \frac{1}{2}mv^2\]

Rearranging for velocity \(v\):
\[v^2 = \frac{2eV}{m}\]
\[v = \sqrt{\frac{2eV}{m}}\]

**(b)**
The magnetic force on the electron acts perpendicular to its direction of motion, providing the centripetal force required for circular motion:
\[F_B = F_c\]
\[Bev = \frac{mv^2}{r}\]

Rearranging to express the charge-to-mass ratio \(\frac{e}{m}\):
\[\frac{e}{m} = \frac{v}{Br}\]

Substitute the expression for \(v\) from part (a):
\[\frac{e}{m} = \frac{1}{Br}\sqrt{\frac{2eV}{m}}\]

Squaring both sides:
\[\left(\frac{e}{m}\right)^2 = \frac{2eV}{m B^2 r^2}\]

Dividing both sides by \(\frac{e}{m}\):
\[\frac{e}{m} = \frac{2V}{B^2 r^2}\]

**(c)**
First, convert the radius from centimeters to meters:
\[r = 5.0\text{ cm} = 0.050\text{ m}\]

Rearrange the derived formula to solve for \(B\):
\[B^2 = \frac{2Vm}{e r^2}\]
\[B = \sqrt{\frac{2Vm}{e r^2}} = \frac{1}{r}\sqrt{\frac{2Vm}{e}}\]

Substitute the given values into the equation:
\[B = \frac{1}{0.050}\sqrt{\frac{2 \times 150 \times 9.11 \times 10^{-31}}{1.60 \times 10^{-19}}}\]
\[B = 20 \times \sqrt{1.708125 \times 10^{-9}}\]
\[B = 20 \times 4.13295 \times 10^{-5}\]
\[B = 8.2659 \times 10^{-4}\text{ T}\]

Rounding to 2 significant figures gives:
\[B = 8.3 \times 10^{-4}\text{ T}\]

**(a) [2 Marks]**
* **M1**: Equating kinetic energy gained to electrical work done: \(eV = \frac{1}{2}mv^2\) (or equivalent in words).
* **A1**: Correct algebraic manipulation leading to the shown equation \(v = \sqrt{\frac{2eV}{m}}\).

**(b) [3 Marks]**
* **M1**: Equating magnetic force to centripetal force: \(Bev = \frac{mv^2}{r}\).
* **M1**: Expressing \(\frac{e}{m} = \frac{v}{Br}\) or substituting the expression for \(v\) from part (a) into a correct force equation.
* **A1**: Fully correct derivation showing \(\frac{e}{m} = \frac{2V}{B^2 r^2}\).

**(c) [4 Marks]**
* **C1**: Conversion of radius to meters: \(r = 0.050\text{ m}\).
* **C1**: Rearrangement of the formula from (b) to make \(B\) the subject: \(B = \sqrt{\frac{2Vm}{e r^2}}\).
* **C1**: Correct substitution of all numeric values into the rearranged formula.
* **A1**: Correct final value with unit: \(B = 8.3 \times 10^{-4}\text{ T}\) (accept \(8.27 \times 10^{-4}\text{ T}\) or \(8.26 \times 10^{-4}\text{ T}\)). Reject negative values. Unit must be T or equivalent (e.g., \(\text{Wb m}^{-2}\)).

The radius of a charged particle's path in a magnetic field is given by \(r = \frac{mv}{Bq}\). Since kinetic energy is \(E_k = \frac{p^2}{2m}\), momentum is \(p = mv = \sqrt{2mE_k}\). Thus, the radius can be written as \(r = \frac{\sqrt{2mE_k}}{Bq}\). For the second particle with mass \(2m\) and charge \(0.5q\), the new radius is \(r' = \frac{\sqrt{2(2m)E_k}}{B(0.5q)} = \frac{\sqrt{2}\sqrt{2mE_k}}{0.5Bq} = 2\sqrt{2} R\).

1 mark for selecting the correct option C.

The r.m.s. speed of an ideal gas is given by \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), where \(m\) is the mass of a single molecule and \(T\) is the absolute temperature. Since the escape of gas does not change the mass of individual molecules, the r.m.s. speed depends solely on the temperature. When the temperature increases from \(T\) to \(3T\), the r.m.s. speed increases by a factor of \\sqrt{3}\), resulting in a new speed of \(\sqrt{3} v\).

1 mark for selecting the correct option B.

After three half-lives, the fraction of active radioactive nuclei remaining is \((1/2)^3 = 1/8\). Therefore, the fraction of nuclei that have decayed is \(1 - 1/8 = 7/8\).

1 mark for selecting the correct option C.

According to Wien's Law, the peak wavelength is inversely proportional to temperature, so \(T \propto 1/\lambda\). Since Star Y has double the peak wavelength of Star X, its temperature is half that of Star X (\(T_Y = 0.5 T_X\)). According to the Stefan-Boltzmann Law, the luminosity is proportional to the fourth power of temperature, \(L \propto R^2 T^4\). Since the radii of both stars are equal, the luminosity of Star Y is \(L_Y = L_X \times (0.5)^4 = L / 16\).

1 mark for selecting the correct option A.

The total energy of a simple harmonic oscillator is \(E_T = \frac{1}{2} k A^2\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\). The kinetic energy is \(E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2)\). Setting \(E_k = E_p\) gives \(\frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k x^2\), which simplifies to \(A^2 - x^2 = x^2\), or \(2x^2 = A^2\). Solving for \(x\) yields \(x = \frac{A}{\sqrt{2}}\).

1 mark for selecting the correct option B.

The centripetal force required to keep the car in circular motion is provided by the static frictional force. Therefore, \(\frac{mv^2}{r} \le \mu m g\). Rearranging this inequality for the speed \(v\) gives \(v \le \sqrt{\mu g r}\). The maximum speed before slipping occurs is thus \(\sqrt{\mu g r}\).

1 mark for selecting the correct option B.

A baryon must consist of three quarks. A strangeness of \(-1\) indicates the presence of exactly one strange quark (\(s\)), which has a charge of \(-1/3\). For the total charge to be \(+1\), the sum of the charges of the other two quarks must be \(+4/3\). Since up quarks (\(u\)) have a charge of \(+2/3\), a combination of two up quarks and one strange quark (\(uus\)) has a total charge of \(+2/3 + 2/3 - 1/3 = +1\) and a strangeness of \(-1\).

1 mark for selecting the correct option A.

The energy stored in a capacitor is proportional to the square of its voltage: \(E(t) = \frac{1}{2} C [V(t)]^2\). During discharge, the voltage varies as \(V(t) = V_0 e^{-t/RC}\). Therefore, the energy at time \(t\) is \(E(t) = E_0 e^{-2t/RC}\). Setting \(E(t) = 0.5 E_0\) gives \(e^{-2t/RC} = 0.5\), which leads to \(-2t/RC = -\ln(2)\). Rearranging for \(t\) gives \(t = \frac{1}{2} RC \ln(2)\).

1 mark for selecting the correct option C.

Using the formula for the radius of a charged particle in a magnetic field:
\(r = \frac{mv}{Bq}\)

Since kinetic energy is \(E_k = \frac{1}{2}mv^2\), the momentum of the particle is \(mv = \sqrt{2mE_k}\).

Substituting this into the radius formula:
\(r = \frac{\sqrt{2mE_k}}{Bq}\)

For particles with the same kinetic energy in the same magnetic field, the radius is proportional to \(\frac{\sqrt{m}}{q}\):
\(r \propto \frac{\sqrt{m}}{q}\)

For a proton: \(m_p = m\) and \(q_p = e\).

For an alpha particle: \(m_{\alpha} = 4m\) and \(q_{\alpha} = 2e\).

Therefore, the ratio of the radii is:
\(\frac{r_{\alpha}}{r_p} = \frac{\sqrt{4m}}{2e} \div \frac{\sqrt{m}}{e} = \frac{2\sqrt{m}}{2e} \times \frac{e}{\sqrt{m}} = 1\)

[1 mark] B - Correct answer.

Incorrect options:
- A: Incorrect ratio from omitting the square root on the mass factor.
- C: Incorrect ratio.
- D: Incorrect ratio from failing to divide by the charge of the alpha particle.

According to Wien's displacement law:
\(\lambda_{\text{max}} T = \text{constant} \implies T \propto \frac{1}{\lambda_{\text{max}}}\)

Given that \(\lambda_X = 2\lambda_Y\), the temperature of Star X is:
\(T_X = \frac{1}{2}T_Y\)

According to the Stefan-Boltzmann law:
\(L = 4\pi R^2 \sigma T^4 \implies R \propto \frac{\sqrt{L}}{T^2}\)

Therefore, the ratio of their radii is:
\(\frac{R_X}{R_Y} = \sqrt{\frac{L_X}{L_Y}} \times \left(\frac{T_Y}{T_X}\right)^2\)

Substitute the known ratios into the equation:
\(\frac{R_X}{R_Y} = \sqrt{4} \times (2)^2 = 2 \times 4 = 8\)

[1 mark] D - Correct answer.

Incorrect options:
- A: Incorrect ratio resulting from inverted temperature calculations.
- B: Incorrect ratio resulting from failing to square the temperature ratio.
- C: Incorrect ratio resulting from omitting the temperature effect completely.

(a) Convert temperature to Kelvin: \( T = 27.0 + 273.15 = 300.15 \text{ K} \) (or use \( 300 \text{ K} \)).
Using the ideal gas equation: \( pV = NkT \)
Rearranging for \( N \):
\( N = \frac{pV}{kT} = \frac{1.20 \times 10^5 \text{ Pa} \times 0.045 \text{ m}^3}{1.38 \times 10^{-23} \text{ J K}^{-1} \times 300.15 \text{ K}} \approx 1.30 \times 10^{24} \text{ atoms} \).

(b) Since the volume of the container is constant:
\( \frac{p_1}{T_1} = \frac{p_2}{T_2} \)
\( T_2 = T_1 \times \frac{p_2}{p_1} = 300.15 \text{ K} \times \frac{1.80 \times 10^5 \text{ Pa}}{1.20 \times 10^5 \text{ Pa}} = 450.2 \text{ K} \approx 450 \text{ K} \).

(c) The kinetic energy of one atom is given by \( E_k = \frac{3}{2} kT \).
The change in total kinetic energy for \( N \) atoms is:
\( \Delta E_k = \frac{3}{2} N k \Delta T = \frac{3}{2} V \Delta p \)
Using \( \Delta E_k = 1.5 \times 0.045 \text{ m}^3 \times (1.80 - 1.20) \times 10^5 \text{ Pa} \):
\( \Delta E_k = 1.5 \times 0.045 \times 60000 = 4050 \text{ J} \).

(a)
- 1 mark: Converts temperature to Kelvin (\( 300 \text{ K} \)).
- 1 mark: Correct substitution into \( pV = NkT \).
- 1 mark: Correct calculation of \( N \) to 2 or 3 sig figs (\( 1.30 \times 10^{24} \)).

(b)
- 1 mark: Identifies constant volume condition and sets up \( p_1/T_1 = p_2/T_2 \).
- 1 mark: Calculates value close to \( 450 \text{ K} \) with work shown.

(c)
- 1 mark: Recognizes total kinetic energy formula \( E_k = 1.5 NkT \) or \( 1.5 pV \).
- 1 mark: Correct substitutions of changes (\( \Delta T = 150 \text{ K} \) or \( \Delta p = 60000 \text{ Pa} \)).
- 1 mark: Correct value (\( 4050 \text{ J} \), accept \( 4100 \text{ J} \)).

(a) The angular frequency is given by:
\( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{14.0 \text{ N m}^{-1}}{0.350 \text{ kg}}} = \sqrt{40} \approx 6.32 \text{ rad s}^{-1} \).
Maximum acceleration is:
\( a_{\text{max}} = \omega^2 A = 40 \text{ rad}^2\text{ s}^{-2} \times 0.080 \text{ m} = 3.20 \text{ m s}^{-2} \).

(b) The velocity at displacement \( x \) is given by:
\( v = \pm \omega \sqrt{A^2 - x^2} \)
\( v = 6.32 \text{ rad s}^{-1} \times \sqrt{0.080^2 - 0.050^2} = 6.32 \times \sqrt{0.0039} = 0.395 \text{ m s}^{-1} \).

(c) The time period of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \).
If mass \( m \) increases while the spring constant \( k \) remains the same, the ratio \( m/k \) increases. Therefore, the period of oscillation \( T \) will increase.

(a)
- 1 mark: Calculates angular frequency squared \( \omega^2 = 40 \text{ s}^{-2} \) (or \( \omega = 6.32 \text{ rad s}^{-1} \)).
- 1 mark: Recalls or uses \( a_{\text{max}} = \omega^2 A \).
- 1 mark: Shows substitution leading to \( 3.2 \text{ m s}^{-2} \).

(b)
- 1 mark: Recalls or uses \( v = \pm \omega \sqrt{A^2 - x^2} \).
- 1 mark: Correct substitution of values (\( A = 0.080 \), \( x = 0.050 \), \( \omega = 6.32 \)).
- 1 mark: Correct velocity calculation (\( 0.395 \text{ m s}^{-1} \) or \( 0.40 \text{ m s}^{-1} \)).

(c)
- 1 mark: States that the period increases.
- 1 mark: References the formula \( T = 2\pi \sqrt{\frac{m}{k}} \) and relates \( T \) to \( \sqrt{m} \).

(a) The activity decreases by a factor of:
\( \frac{A}{A_0} = \frac{2.1 \times 10^5}{8.4 \times 10^5} = 0.25 = \left(\frac{1}{2}\right)^2 \).
This corresponds to exactly two half-lives.
\( 2 \times T_{1/2} = 15.0 \text{ hours} \implies T_{1/2} = 7.5 \text{ hours} \).

(b) Convert half-life to seconds:
\( T_{1/2} = 7.5 \text{ hours} \times 3600 \text{ s hour}^{-1} = 27000 \text{ s} \).
Use the relationship between half-life and decay constant:
\( \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{27000 \text{ s}} \approx 2.57 \times 10^{-5} \text{ s}^{-1} \).

(c) Calculate the remaining activity \( A \) after \( t = 24 \text{ hours} = 86400 \text{ s} \):
\( A = A_0 e^{-\lambda t} = 8.4 \times 10^5 \text{ Bq} \times e^{-(2.57 \times 10^{-5} \times 86400)} \approx 9.12 \times 10^4 \text{ Bq} \).
Using the relationship \( A = \lambda N \):
\( N = \frac{A}{\lambda} = \frac{9.12 \times 10^4 \text{ Bq}}{2.57 \times 10^{-5} \text{ s}^{-1}} \approx 3.55 \times 10^9 \text{ nuclei} \).

(a)
- 1 mark: Shows that the activity has halved twice (reduced to 1/4 of initial value).
- 1 mark: Deduces \( T_{1/2} = 15.0 / 2 = 7.5 \text{ hours} \).

(b)
- 1 mark: Correct conversion of hours to seconds (\( 27000 \text{ s} \)).
- 1 mark: Uses \( \lambda = \frac{\ln 2}{T_{1/2}} \).
- 1 mark: Correct calculation of decay constant (\( 2.57 \times 10^{-5} \text{ s}^{-1} \) or \( 2.6 \times 10^{-5} \text{ s}^{-1} \)).

(c)
- 1 mark: Calculates the remaining activity after 24 hours (\( 9.12 \times 10^4 \text{ Bq} \) to \( 9.20 \times 10^4 \text{ Bq} \)) or equivalent fraction method.
- 1 mark: Recalls or uses \( A = \lambda N \).
- 1 mark: Calculates \( N \) correctly to 2 or 3 sig figs (\( 3.5 \times 10^9 \) to \( 3.6 \times 10^9 \)).

(a) Wien's Displacement Law:
\( \lambda_{\text{max}} T = 2.898 \times 10^{-3} \text{ m K} \)
\( T = \frac{2.898 \times 10^{-3} \text{ m K}}{410 \times 10^{-9} \text{ m}} \approx 7068 \text{ K} \approx 7070 \text{ K} \).

(b) Stefan-Boltzmann Law:
\( L = 4\pi R^2 \sigma T^4 \)
Rearranging for radius \( R \):
\( R = \sqrt{\frac{L}{4\pi \sigma T^4}} = \sqrt{\frac{4.5 \times 10^{27} \text{ W}}{4\pi \times 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4} \times (7068 \text{ K})^4}} \)
\( R = \sqrt{\frac{4.5 \times 10^{27}}{1.78 \times 10^9}} \approx 1.59 \times 10^9 \text{ m} \).

(c) Redshift equation:
\( z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \)
\( \Delta \lambda = 658.2 \text{ nm} - 656.3 \text{ nm} = 1.9 \text{ nm} \)
\( v = c \times \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8 \text{ m s}^{-1} \times \frac{1.9 \text{ nm}}{656.3 \text{ nm}} \approx 8.68 \times 10^5 \text{ m s}^{-1} \).

(a)
- 1 mark: Recalls or uses Wien's Displacement Law formula.
- 1 mark: Calculates temperature (\( 7070 \text{ K} \) or \( 7100 \text{ K} \)).

(b)
- 1 mark: Recalls or uses \( L = 4\pi R^2 \sigma T^4 \).
- 1 mark: Correct algebraic rearrangement to isolate \( R \).
- 1 mark: Calculates radius (\( 1.59 \times 10^9 \text{ m} \) to \( 1.62 \times 10^9 \text{ m} \) depending on rounding of \( T \)).

(c)
- 1 mark: Calculates change in wavelength (\( 1.9 \text{ nm} \)).
- 1 mark: Recalls or uses redshift velocity formula \( v/c = \Delta \lambda / \lambda \).
- 1 mark: Correctly calculates \( v = 8.68 \times 10^5 \text{ m s}^{-1} \) (or \( 8.7 \times 10^5 \text{ m s}^{-1} \)).

(a) The magnetic force acts as the centripetal force:
\( B q v = \frac{m v^2}{r} \)
Rearranging for speed \( v \):
\( v = \frac{B q r}{m} \)
Substitute values where \( q = 1.60 \times 10^{-19} \text{ C} \):
\( v = \frac{0.45 \text{ T} \times 1.60 \times 10^{-19} \text{ C} \times 0.12 \text{ m}}{3.20 \times 10^{-26} \text{ kg}} = 2.70 \times 10^5 \text{ m s}^{-1} \).

(b) The period \( T \) of one full orbit is:
\( T = \frac{2\pi r}{v} \)
\( T = \frac{2\pi \times 0.12 \text{ m}}{2.70 \times 10^5 \text{ m s}^{-1}} \approx 2.79 \times 10^{-6} \text{ s} \).

(c) Since \( r = \frac{m v}{B q} \), path radius \( r \) is inversely proportional to the charge \( q \) (since mass, speed, and magnetic field are constant). Double charge means the radius is halved, i.e., it becomes \( 0.060 \text{ m} \).

(a)
- 1 mark: Equates magnetic force \( Bqv \) to centripetal force \( mv^2/r \).
- 1 mark: Correct rearrangement to isolate speed \( v \).
- 1 mark: Shows calculation leading to \( 2.70 \times 10^5 \text{ m s}^{-1} \).

(b)
- 1 mark: Recalls or uses \( T = 2\pi r / v \).
- 1 mark: Substitutes values correctly.
- 1 mark: Obtains \( 2.79 \times 10^{-6} \text{ s} \) (accept \( 2.8 \times 10^{-6} \text{ s} \)).

(c)
- 1 mark: States that the radius is halved (or reduced to \( 0.060 \text{ m} \)).
- 1 mark: Explains that radius is inversely proportional to charge (\( r \propto 1/q \)).

(a) Using the de Broglie equation:
\( \lambda = \frac{h}{p} \implies p = \frac{h}{\lambda} \)
\( p = \frac{6.63 \times 10^{-34} \text{ J s}}{1.5 \times 10^{-10} \text{ m}} = 4.42 \times 10^{-24} \text{ N s} \).

(b) Kinetic energy gained is given by:
\( E_k = \frac{p^2}{2m_e} = e V \)
Rearranging for accelerating voltage \( V \):
\( V = \frac{p^2}{2 m_e e} = \frac{(4.42 \times 10^{-24} \text{ N s})^2}{2 \times 9.11 \times 10^{-31} \text{ kg} \times 1.60 \times 10^{-19} \text{ C}} \)
\( V = \frac{1.954 \times 10^{-47}}{2.915 \times 10^{-49}} \approx 67.0 \text{ V} \).

(c) When high-energy electrons collide with a proton (deep inelastic scattering):
- Electrons are observed to scatter through much larger angles than expected if the proton's charge was uniform.
- This pattern is characteristic of scattering from small, point-like, charged objects within the proton.
- This provides experimental evidence for the existence of quarks as the constituents of protons.

(a)
- 1 mark: Recalls or uses \( p = h/\lambda \).
- 1 mark: Calculates momentum as \( 4.42 \times 10^{-24} \text{ N s} \).

(b)
- 1 mark: Equates electrical potential energy lost \( eV \) to kinetic energy \( p^2 / 2m_e \).
- 1 mark: Correct substitution of values.
- 1 mark: Correct calculation of voltage \( 67.0 \text{ V} \) (accept \( 66.5 \text{ V} \) to \( 67.5 \text{ V} \)).

(c)
- 1 mark: Mentions that electrons scatter through surprisingly large angles.
- 1 mark: Explains that this implies charge is not uniformly distributed but concentrated in point-like centers.
- 1 mark: Concludes that this is evidence for internal structure (quarks).

(a) Centripetal acceleration:
\( a_c = \frac{v^2}{r} = \frac{(2.50 \text{ m s}^{-1})^2}{0.45 \text{ m}} \approx 13.9 \text{ m s}^{-2} \).

(b) At the top of the loop:
- The weight (\( W = mg \)) acts vertically downwards.
- The normal contact force (\( N \)) exerted by the track on the car also acts vertically downwards (towards the center of the circular path).
Both vectors must point downwards.

(c) The net downward force provides the centripetal force:
\( F_{\text{net}} = W + N = \frac{m v^2}{r} \)
\( m g + N = m a_c \implies N = m(a_c - g) \)
\( N = 0.120 \text{ kg} \times (13.89 \text{ m s}^{-2} - 9.81 \text{ m s}^{-2}) = 0.120 \times 4.08 \approx 0.49 \text{ N} \).

(d) The car stays in contact with the track if \( N \ge 0 \).
The limiting condition is when \( N = 0 \):
\( m g = \frac{m v_{\text{min}}^2}{r} \implies v_{\text{min}} = \sqrt{g r} = \sqrt{9.81 \text{ m s}^{-2} \times 0.45 \text{ m}} \approx 2.10 \text{ m s}^{-1} \).

(a)
- 1 mark: Recalls or uses \( a_c = v^2/r \).
- 1 mark: Correctly calculates \( 13.9 \text{ m s}^{-2} \).

(b)
- 1 mark: Draws weight vector pointing vertically downwards.
- 1 mark: Draws normal contact force vector pointing vertically downwards.

(c)
- 1 mark: Correctly sets up equation for centripetal force: \( N + mg = mv^2/r \).
- 1 mark: Correct substitution of values.
- 1 mark: Correctly calculates normal force as \( 0.49 \text{ N} \).

(d)
- 1 mark: Correct calculation of minimum speed (\( 2.10 \text{ m s}^{-1} \)).

(a) Convert the parallax angle to parsecs:
\( d \text{ (in pc)} = \frac{1}{p \text{ (in arcseconds)}} = \frac{1}{0.185} \approx 5.405 \text{ pc} \).
Convert parsecs to meters:
\( d = 5.405 \text{ pc} \times 3.09 \times 10^{16} \text{ m pc}^{-1} \approx 1.67 \times 10^{17} \text{ m} \).
(Alternatively, using trigonometry: \( \theta = \frac{0.185}{3600} = 5.139 \times 10^{-5} \text{ degrees} \approx 8.97 \times 10^{-7} \text{ radians} \). Using the Earth-Sun distance \( r = 1.50 \times 10^{11} \text{ m} \), \( d = \frac{r}{\tan\theta} = \frac{1.50 \times 10^{11}}{\tan(8.97 \times 10^{-7})} \approx 1.67 \times 10^{17} \text{ m} \)).

(b) As the distance to a star increases, the parallax angle decreases. For very distant stars (the vast majority of stars in our galaxy), the angle becomes too small to measure with sufficient precision due to limits in angular resolution and atmospheric interference.

(c) A star like the Sun (low-mass) will expand into a red giant, shed its outer layers as a planetary nebula, and leave behind a dense, stable white dwarf core.
A high-mass star (15 solar masses) will expand into a red supergiant, undergo a violent supernova explosion, and leave behind either a neutron star or a black hole.

(a)
- 1 mark: Calculates distance in parsecs (\( 5.41 \text{ pc} \)) or angle in radians (\( 8.97 \times 10^{-7} \text{ rad} \)).
- 1 mark: Uses correct conversion factor (\( 3.09 \times 10^{16} \text{ m} \)) or trigonometric relation.
- 1 mark: Obtains value close to \( 1.67 \times 10^{17} \text{ m} \) with units.

(b)
- 1 mark: Identifies that the parallax angle decreases as distance increases.
- 1 mark: Explains that the angle becomes too small to measure/resolve with existing instruments.

(c)
- 1 mark: Low-mass star / Sun evolves into a red giant then ends as a white dwarf.
- 1 mark: High-mass star evolves into a red supergiant, then undergoes a supernova.
- 1 mark: High-mass star remnant is a neutron star or black hole.

Part (a):
Convert temperature to Kelvin: \( T = 20 + 273.15 = 293.15 \text{ K} \) (or \( 293 \text{ K} \)).
Using the r.m.s. speed formula: \( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \)
Substitute values: \( v_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 293}{4.00 \times 10^{-3}}} = \sqrt{1.826 \times 10^6} \approx 1350 \text{ m s}^{-1} \).

Part (b):
1. Higher temperature increases the mean kinetic energy of the helium atoms, increasing their mean speed.
2. This results in a greater change in momentum during each collision with the canister walls.
3. Faster atoms also collide with the walls more frequently.
4. By Newton's second law (\( F = \frac{\Delta p}{\Delta t} \)), the greater momentum change and higher frequency of collisions increase the average force on the walls.
5. Since \( p = \frac{F}{A} \) and area is constant, pressure increases.

Part (a) [3 Marks]:
- Correct conversion of temperature to Kelvin: \( T = 293 \text{ K} \) (1)
- Correct formula substitution: \( v_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 293}{4.00 \times 10^{-3}}} \) (or equivalent using single-atom mass) (1)
- Correct final answer: \( v_{\text{rms}} = 1350 \text{ m s}^{-1} \) (accept \( 1351 \text{ m s}^{-1} \)) (1)

Part (b) [5 Marks]:
- Identifies that increased temperature increases kinetic energy and mean speed of atoms (1)
- Identifies that this leads to a larger change in momentum during collisions (1)
- Identifies that there is a higher frequency of collisions (1)
- Explains that the average force increases due to rate of change of momentum (1)
- Relates increased force to increased pressure (1)

Part (a):
Electrical work done equals kinetic energy: \( qV = \frac{1}{2}mv^2 \)
\( v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times (3.20 \times 10^{-19} \text{ C}) \times 12000 \text{ V}}{6.64 \times 10^{-27} \text{ kg}}} = \sqrt{1.157 \times 10^{12}} \approx 1.075 \times 10^6 \text{ m s}^{-1} \approx 1.1 \times 10^6 \text{ m s}^{-1} \).

Using centripetal force from magnetic field: \( Bqv = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{Bq} \)
\( r = \frac{(6.64 \times 10^{-27} \text{ kg}) \times (1.075 \times 10^6 \text{ m s}^{-1})}{0.45 \text{ T} \times (3.20 \times 10^{-19} \text{ C})} \approx 0.0496 \text{ m} \approx 0.050 \text{ m} \).

Part (b):
Using Fleming's Left-Hand Rule, the magnetic force is upwards. For no deflection, the electric force must be downwards. Because the alpha particle is positive, the electric field direction must also be downwards (towards the bottom of the page).
For balanced forces: \( qE = Bqv \Rightarrow E = Bv \)
\( E = 0.45 \text{ T} \times 1.075 \times 10^6 \text{ m s}^{-1} \approx 4.8 \times 10^5 \text{ V m}^{-1} \) (or \( 4.95 \times 10^5 \text{ V m}^{-1} \) if using the rounded \( 1.1 \times 10^6 \text{ m s}^{-1} \)).

Part (a) [4 Marks]:
- Equates electrical energy to kinetic energy: \( qV = \frac{1}{2}mv^2 \) (1)
- Shows calculation leading to \( v \approx 1.1 \times 10^6 \text{ m s}^{-1} \) (1)
- Equates magnetic force to centripetal force: \( Bqv = \frac{mv^2}{r} \) (1)
- Calculates radius \( r = 0.050 \text{ m} \) (accept range \( 0.049 \text{ m} \) to \( 0.051 \text{ m} \)) (1)

Part (b) [4 Marks]:
- Identifies magnetic force is upwards using Fleming's Left-Hand Rule (1)
- States electric field must point downwards (towards the bottom of the page) (1)
- Uses condition for no deflection: \( E = Bv \) (1)
- Calculates electric field magnitude in range \( 4.8 \times 10^5 \text{ V m}^{-1} \) to \( 5.0 \times 10^5 \text{ V m}^{-1} \) (1)

(a) Taking the natural logarithm of both sides of the discharge equation:
\(\ln V = \ln(V_0 e^{-\frac{t}{RC}})\)
\(\ln V = \ln V_0 + \ln(e^{-\frac{t}{RC}})\)
\(\ln V = -\frac{1}{RC} t + \ln V_0\)
Comparing this to the equation of a straight line, \(y = mx + c\):
- \(y = \ln V\)
- \(x = t\)
- \(m = -\frac{1}{RC}\)
- \(c = \ln V_0\)
Therefore, a graph of \(\ln V\) against \(t\) yields a straight line with gradient \(-\frac{1}{RC}\).

(b) For \(t = 45.0 \text{ s}\), \(V = 3.47 \text{ V}\).
\(\ln(3.47) = 1.244\) (or \(1.24\)).

(c)(i) Given gradient \(m = -0.0212 \text{ s}^{-1}\):
\(m = -\frac{1}{RC} \implies C = -\frac{1}{R m}\)
\(C = \frac{1}{150 \times 10^3 \text{ } \Omega \times 0.0212 \text{ s}^{-1}} = \frac{1}{3180} \approx 3.145 \times 10^{-4} \text{ F}\) (or \(314 \text{ } \mu\text{F}\)).

(c)(ii) Percentage uncertainty in gradient = \(\frac{0.0004}{0.0212} \times 100\% \approx 1.89\%\).

(c)(iii) Percentage uncertainty in \(R\) = \(\frac{3 \text{ k}\Omega}{150 \text{ k}\Omega} \times 100\% = 2.00\%\).

(c)(iv) Since \(C = \frac{1}{R |m|}\):
\(\% \text{ uncertainty in } C = \% \text{ uncertainty in } R + \% \text{ uncertainty in } |m|\)
\(\% \text{ uncertainty in } C = 2.00\% + 1.89\% = 3.89\%\)
Absolute uncertainty in \(C = 3.89\% \times 3.145 \times 10^{-4} \text{ F} \approx 1.22 \times 10^{-5} \text{ F}\) (or \(12 \text{ } \mu\text{F}\)).

(d) Modification: Use a larger resistor \(R\) (or a larger capacitor \(C\)) to increase the time constant \(\tau = RC\).
Explanation: This makes the discharge process slower, so the total time measured is larger. Consequently, the human reaction time (about \(0.2\) s) becomes a smaller fraction of the total time, reducing the percentage uncertainty.
Alternative Modification: Use a data logger with a voltage sensor instead of manual stopwatch recording.
Alternative Explanation: This eliminates human reaction time entirely, dramatically reducing the uncertainty in time measurements.

- (a) [2 Marks]
- 1 Mark: Correctly applies natural logs to the equation to get \(\ln V = \ln V_0 - \frac{t}{RC}\).
- 1 Mark: Compares to \(y = mx + c\) and clearly identifies gradient \(m = -1/RC\).
- (b) [1 Mark]
- 1 Mark: Correctly calculates \(\ln(3.47) = 1.24\) (allow 1.24 or 1.244).
- (c)(i) [3 Marks]
- 1 Mark: Uses \(C = 1 / (R \times |\text{gradient}|)\).
- 1 Mark: Substitutes correct values including powers of ten: \(150 \times 10^3\).
- 1 Mark: Correct final value of \(3.14 \times 10^{-4} \text{ F}\) (accept \(314 \text{ } \mu\text{F}\) to \(3.15 \times 10^{-4} \text{ F}\)).
- (c)(ii) [1 Mark]
- 1 Mark: Correctly calculates \(1.89\%\) (accept \(1.9\%\)).
- (c)(iii) [1 Mark]
- 1 Mark: Correctly calculates \(2.00\%\) (accept \(2\%\)).
- (c)(iv) [2.5 Marks]
- 1 Mark: States that percentage uncertainties are added for a quotient/product: \(\% \text{unc} = \% \Delta R + \% \Delta m\).
- 0.5 Mark: Obtains total percentage uncertainty of \(3.89\%\) (or \(3.9\%\)).
- 1 Mark: Correctly calculates absolute uncertainty as \(1.2 \times 10^{-5} \text{ F}\) (or \(12 \text{ } \mu\text{F}\)), rounded to 1 or 2 sig figs with correct units.
- (d) [2 Marks]
- 1 Mark: Identifies a valid modification (e.g., larger resistance, larger capacitor, or using a data logger).
- 1 Mark: Provides a corresponding correct explanation (e.g., increases discharge time to reduce relative reaction time, or eliminates reaction time).

(a) Starting with \(P t = m c (\theta - \theta_0)\):
\(\theta - \theta_0 = \frac{P}{mc} t\)
\(\theta = \left(\frac{P}{mc}\right) t + \theta_0\)
Comparing this to the equation of a straight line, \(y = mx + c\):
- \(y = \theta\)
- \(x = t\)
- \(m_g = \frac{P}{mc}\) (gradient)
- \(c_{\text{intercept}} = \theta_0\)
Since \(P\), \(m\), and \(c\) are constant, the gradient is constant, indicating a straight-line relationship.

(b)(i) Power is calculated by:
\(P = V I = 11.5 \text{ V} \times 4.20 \text{ A} = 48.3 \text{ W}\) (or \(48.30 \text{ W}\)).

(b)(ii) Since the gradient \(m_g = \frac{P}{mc}\):
\(c = \frac{P}{m \cdot m_g}\)
\(c = \frac{48.3 \text{ W}}{1.02 \text{ kg} \times 0.0485 \text{ }^{\circ}\text{C s}^{-1}} = \frac{48.3}{0.04947} \approx 976.35 \text{ J kg}^{-1} \text{ K}^{-1}\) (or \(\text{J kg}^{-1} \text{ }^{\circ}\text{C}^{-1}\)).
We express this to 3 significant figures as \(976 \text{ J kg}^{-1} \text{ K}^{-1}\).

(b)(iii) For power \(P = V I\), the percentage uncertainties add:
\(\% \Delta P = \% \Delta V + \% \Delta I\)
\(\% \Delta V = \frac{0.2}{11.5} \times 100\% = 1.74\%\)
\(\% \Delta I = \frac{0.05}{4.20} \times 100\% = 1.19\%\)
\(\% \Delta P = 1.74\% + 1.19\% = 2.93\%\) (accept \(2.9\%\)).

(b)(iv) Since \(c = \frac{P}{m \cdot m_g}\) and the uncertainty in \(m_g\) is negligible:
\(\% \Delta c = \% \Delta P + \% \Delta m\)
\(\% \Delta m = \frac{0.01}{1.02} \times 100\% = 0.98\%\)
\(\% \Delta c = 2.93\% + 0.98\% = 3.91\%\) (accept \(3.9\%\)).

(c) Two practical measures to minimize heat loss:
1. Wrap the aluminum block in an insulating material (such as cotton wool, felt, or bubble wrap) to reduce conduction and convection to the surrounding air.
2. Place the block on a heat-resistant insulating mat (e.g., polystyrene or cork board) rather than directly on a metal bench to prevent conduction downwards.
3. Put a small drop of oil or glycerol in the thermometer/heater holes to improve thermal contact and reduce temperature differences.

- (a) [2 Marks]
- 1 Mark: Correctly rearranges the formula to \(\theta = \frac{P}{mc} t + \theta_0\).
- 1 Mark: States that gradient is \(\frac{P}{mc}\) which is a constant, hence giving a straight line of form \(y = mx + c\).
- (b)(i) [1 Mark]
- 1 Mark: Calculates power correctly as \(48.3 \text{ W}\).
- (b)(ii) [3 Marks]
- 1 Mark: Equates gradient to \(\frac{P}{mc}\) and rearranges for \(c\).
- 1 Mark: Substitutes correct values into the formula.
- 1 Mark: Obtains \(976 \text{ J kg}^{-1} \text{ K}^{-1}\) (accept \(970\) to \(980 \text{ J kg}^{-1} \text{ K}^{-1}\) depending on rounding of intermediate values; must have correct unit).
- (b)(iii) [2 Marks]
- 1 Mark: Adds percentage uncertainties of \(V\) and \(I\).
- 1 Mark: Obtains \(2.93\%\) (accept \(2.9\%\)).
- (b)(iv) [2.5 Marks]
- 1 Mark: Realizes that \(\% \Delta c = \% \Delta P + \% \Delta m\).
- 0.5 Mark: Correctly calculates \(\% \Delta m = 0.98\%\).
- 1 Mark: Obtains a final percentage uncertainty of \(3.91\%\) (accept \(3.9\%\)).
- (c) [2 Marks]
- 1 Mark: Suggests insulating the block (e.g., wrapping in cotton wool/felt).
- 1 Mark: Suggests putting oil in the thermometer hole to ensure good thermal contact, or placing the block on an insulating tile.

(a) Measuring 20 oscillations instead of 1 reduces the effect of human reaction time on the measured period. The absolute uncertainty in the total measured time (due to reaction time, typically \(\pm 0.2\) s) is divided by 20 when calculating the period, thereby significantly reducing the percentage uncertainty in \(T\).

(b)(i) First, calculate the mean time for 20 oscillations:
\(t_{\text{mean}} = \frac{18.42 + 18.36 + 18.48}{3} = 18.42 \text{ s}\)
The range of the values is \(18.48 - 18.36 = 0.12 \text{ s}\).
The uncertainty in the total time is half the range:
\(\Delta t = \frac{0.12}{2} = 0.06 \text{ s}\).
Now, calculate the mean period \(T\):
\(T = \frac{18.42}{20} = 0.921 \text{ s}\)
The absolute uncertainty in the period \(T\) is:
\(\Delta T = \frac{0.06}{20} = 0.003 \text{ s}\).
So, \(T = 0.921 \pm 0.003 \text{ s}\).

(b)(ii) The percentage uncertainty in \(T\) is:
\(\% \Delta T = \frac{0.003}{0.921} \times 100\% \approx 0.326\%\)
Since \(T^2 = T \times T\), the percentage uncertainty in \(T^2\) is twice that of \(T\):
\(\% \Delta (T^2) = 2 \times \% \Delta T = 2 \times 0.326\% = 0.65\%\) (accept \(0.65\%\) to \(0.7\%\)).

(c)(i) Squaring both sides of the time period equation:
\(T^2 = 4\pi^2 \frac{m}{k} = \left(\frac{4\pi^2}{k}\right) m\)
Comparing with \(y = mx\), the gradient of a graph of \(T^2\) against \(m\) is \(\frac{4\pi^2}{k}\).
Given gradient \(m_g = 3.95 \text{ s}^2 \text{ kg}^{-1}\):
\(3.95 = \frac{4\pi^2}{k} \implies k = \frac{4\pi^2}{3.95} \approx 9.99 \text{ N m}^{-1}\) (or \(\text{kg s}^{-2}\)).

(c)(ii) Setup details:
1. A retort stand is securely clamped to the bench using a G-clamp to prevent it from tipping over when the mass oscillates (safety).
2. A horizontal rod/clamp holds the spring. A fiducial marker (e.g., a pin or pointer on another retort stand) is placed at the equilibrium position (center of oscillation) to provide a clear reference point for counting complete cycles.
3. A ruler is positioned vertically nearby if displacement amplitude needs monitoring, ensuring the spring is not stretched beyond its elastic limit (safety/accuracy).
4. Oscillations are kept small and strictly vertical to prevent horizontal swaying which changes the motion dynamics.

- (a) [2 Marks]
- 1 Mark: States that it reduces the percentage/relative uncertainty in the time period.
- 1 Mark: Explains that human reaction time is divided by 20 (or spread over 20 oscillations).
- (b)(i) [3 Marks]
- 1 Mark: Calculates mean period correctly as \(0.921 \text{ s}\).
- 1 Mark: Uses the half-range method to find uncertainty in total time as \(0.06 \text{ s}\).
- 1 Mark: Divides uncertainty by 20 to get \(\pm 0.003 \text{ s}\) (must have correct units and format).
- (b)(ii) [2 Marks]
- 1 Mark: Correctly calculates percentage uncertainty of \(T\) as \(0.326\%\).
- 1 Mark: Multiplies by 2 to find percentage uncertainty of \(T^2\) as \(0.65\%\) (accept \(0.65\%\) to \(0.7\%\)).
- (c)(i) [2.5 Marks]
- 1 Mark: Shows the derivation \(T^2 = \frac{4\pi^2}{k} m\) and identifies gradient.
- 1 Mark: Equates gradient to \(3.95\) and rearranges to solve for \(k\).
- 0.5 Mark: Correct final value of \(9.99 \text{ N m}^{-1}\) (accept \(10.0 \text{ N m}^{-1}\)).
- (c)(ii) [3 Marks]
- 1 Mark: Mentions securing retort stand using a G-clamp (safety).
- 1 Mark: Mentions placing a fiducial marker at the equilibrium position to improve counting accuracy (accuracy).
- 1 Mark: Mentions avoiding large amplitudes to prevent exceeding the elastic limit of the spring (safety/accuracy).

(a) Background radiation is present everywhere from natural sources (e.g., cosmic rays, rocks). This background count rate must be subtracted from the total count rate to find the actual (corrected) count rate due solely to the gamma source. To ensure accuracy, the background count should be measured for a long period (e.g., at least 10 minutes) with the source kept far away in its lead container. A long count time reduces the statistical uncertainty of the random decay process when converting to a count rate.

(b)(i) Corrected count rate:
\(R = \text{Total count rate} - \text{Background count rate}\)
\(R = 342 \text{ cpm} - 24 \text{ cpm} = 318 \text{ cpm}\).

(b)(ii) Taking the natural logarithm of both sides:
\(\ln R = \ln(R_0 e^{-\mu x})\)
\(\ln R = \ln R_0 + \ln(e^{-\mu x})\)
\(\ln R = -\mu x + \ln R_0\)
Comparing this to the equation of a straight line, \(y = mx + c\):
- \(y = \ln R\)
- \(x = x\)
- \(m = -\mu\)
- \(c = \ln R_0\)
Since \(\mu\) and \(R_0\) are constants, this represents a straight line with a constant negative gradient of \(-\mu\).

(b)(iii) Substituting the given values into the equation:
\(R = 620 \times e^{-0.083 \times 8.0} = 620 \times e^{-0.664}\)
\(R = 620 \times 0.5148 \approx 319.17 \text{ cpm}\) (accept \(319 \text{ cpm}\)).

(b)(iv) By definition of half-value thickness \(x_{1/2}\):
\(R = 0.5 R_0 \implies 0.5 R_0 = R_0 e^{-\mu x_{1/2}}\)
\(0.5 = e^{-\mu x_{1/2}}\)
Taking natural logs:
\(\ln(0.5) = -\mu x_{1/2} \implies -0.693 = -\mu x_{1/2}\)
\(x_{1/2} = \frac{\ln 2}{\mu} = \frac{0.693}{0.083 \text{ mm}^{-1}} \approx 8.35 \text{ mm}\).

(c) Three safety precautions:
1. Always handle the source using long-handled tongs to maximize the distance between the source and the body, reducing radiation dose (inverse-square law).
2. Point the source aperture away from anyone in the room at all times, ensuring the beam is directed safely towards the GM tube and absorber.
3. Keep the source inside its protective lead-lined storage container when not actively taking measurements to minimize exposure time.

- (a) [3 Marks]
- 1 Mark: Explains that background radiation must be subtracted to isolate the source's contribution.
- 1 Mark: Describes measuring background rate with the source absent (or shielded in lead).
- 1 Mark: Mentions taking the measurement over a long duration (e.g., several minutes) to minimize statistical/random variation.
- (b)(i) [1 Mark]
- 1 Mark: Correctly calculates corrected count rate as \(318 \text{ cpm}\).
- (b)(ii) [2 Marks]
- 1 Mark: Correctly applies natural logarithms to show \(\ln R = -\mu x + \ln R_0\).
- 1 Mark: Relates terms to \(y = mx + c\), showing gradient is \(-\mu\).
- (b)(iii) [1.5 Marks]
- 1 Mark: Substitutes correct numbers into the exponential decay equation.
- 0.5 Mark: Correctly calculates \(319 \text{ cpm}\) (accept \(319\) to \(320\)).
- (b)(iv) [2 Marks]
- 1 Mark: Uses \(x_{1/2} = \frac{\ln 2}{\mu}\).
- 1 Mark: Correctly calculates \(8.35 \text{ mm}\) (accept \(8.3\) to \(8.4\)).
- (c) [3 Marks]
- 1 Mark: Handle with tongs / keep distance.
- 1 Mark: Store in a lead container immediately when not in use.
- 1 Mark: Point away from eyes/body/others.