OCR A Physics OCR Paper 1 2025

  • Section A
  • You should spend a maximum of 30 minutes on this section.
  • 1. Write your answer to each question in the box provided.
  • Which component of a gamma camera absorbs gamma rays and produces visible light? (1)
  • A. collimator
  • B. computer
  • C. photomultiplier tube
  • D. scintillator
  • Result: D
  • Explanation:
  • The scintillator crystal (usually sodium iodide) absorbs incoming gamma rays and converts their energy into visible light photons.
  • These light photons are then detected and amplified by photomultiplier tubes.
  • 2. What is a reasonable estimate for the diameter of an atom? (1)
  • A. [math]10^{-15}\mathrm{m}[/math]
  • B. [math]10^{-12}\mathrm{m}[/math]
  • C. [math]10^{-10}\mathrm{m}[/math]
  • D. [math]10^{-7}\mathrm{m}[/math]
  • Result: C
  • Explanation:
  • Typical atomic diameters are about 0.1 nm.
  • [math]0.1\mathrm{nm}=1\times10^{-10}\mathrm{m}[/math]
  • Therefore [math]10^{-10}\mathrm{m}[/math] is the best estimate.
  • 3. The graph below shows the binding energy per nucleon, E, for nuclei of different nucleon number, A.
  • Which of the following statement(s) correctly describe nuclear processes? (1)
  • 1- Fission of nuclei to the right of X releases energy
  • 2- Fusion of nuclei to the left of X releases energy
  • 3- Fission of nuclei to the right of X can happen spontaneously
  • A. Only 1
  • B. Only 2
  • C. Only 2 and 3
  • D. 1, 2 and 3
  • Result: D
  • Explanation:
  • – Nuclei to the right of X are heavy nuclei.They release energy by fission.
  • – Nuclei to the left of X are light nuclei.They release energy by fusion.
  • – Some very heavy nuclei can undergo spontaneous fission.
  • Therefore, all three statements are correct.
  • The current in a copper wire of radius [math]2.5\times10^{-4}\mathrm{m}[/math] is [math]1.4\mathrm{A}[/math]. (1)
  • The number density of charge carriers (electrons) in copper is [math]8.5\times10^{28}\mathrm{m^{-3}}[/math].
  • 4. What is the mean drift velocity of the electrons in the wire?
  • A. [math]2.1\times10^{-4}\mathrm{mms^{-1}}[/math]
  • B. [math]0.13\mathrm{mms^{-1}}[/math]
  • C. [math]0.52\mathrm{mms^{-1}}[/math]
  • D. [math]1.9\mathrm{mms^{-1}}[/math]
  • Result: C
  • Explanation:
  • Mean Drift Velocity of Electrons:
  • Given data:
  • [math]I=1.4\mathrm{A}[/math]
  • [math]r=2.5\times10^{-4}\mathrm{m}[/math]
  • [math]n=8.5\times10^{28}\mathrm{m^{-3}}[/math]
  • [math]e=1.6\times10^{-19}\mathrm{C}[/math]
  • Using formula:
  • [math]I=nAev_d[/math]
  • Area:
  • [math]A=\pi r^2[/math]
  • [math]A=\pi(2.5\times10^{-4})^2[/math]
  • [math]A=1.96\times10^{-7},\mathrm{m^2}[/math]
  • [math]v_d=\frac{I}{nAe}[/math]
  • [math]v_d=\frac{1.4}{(8.5\times10^{28})(1.96\times10^{-7})(1.6\times10^{-19})}[/math]
  • [math]v_d=5.2\times10^{-4}\mathrm{ms^{-1}}[/math]
  • [math]v_d=0.52\mathrm{mms^{-1}}[/math]
  • 5. An isotope has a half-life of 243 years. (1)
  • One sample of this isotope has activity [math]A_1[/math].
  • A second sample of the same isotope has activity [math]A_2[/math].
  • What is the ratio = [math]\frac{\text{mass of isotope in first sample}}{\text{mass of isotope in second sample}}=?[/math]
  • A. [math]\frac{243A_1}{A_2}[/math]
  • B. [math]\frac{A_1}{243}[/math]
  • C. [math]\frac{A_1}{A_2}[/math]
  • D. [math]\frac{A_2}{243}[/math]
  • Result: C
  • Explanation:
  • [math]\frac{\text{mass of first sample}}{\text{mass of second sample}}=\frac{A_1}{A_2}[/math]
  • Solution:
  • For the same isotope:
  • [math]A=\lambda N[/math]
  • Since [math]\lambda[/math] is constant,
  • [math]A\propto N[/math]
  • And
  • [math]N\propto\text{mass}[/math]
  • Therefore,
  • [math]\frac{m_1}{m_2}=\frac{A_1}{A_2}[/math]
  • 6. A student wants to determine the count-rate from a beta-emitting source in the laboratory. Which procedure would be the best to determine the count-rate? (1)
  • A. Measure the counts over a long period of time
  • B. Place the radiation detector (Geiger tube) about 1 metre away from the source
  • C. Place the source close to a very strong magnet
  • D. Place the source inside a lead box when measuring the counts
  • Result: A
  • Explanation:
  • Radioactive decay is random. Measuring for a long time reduces percentage uncertainty and gives a more accurate count rate.
  • 7. A cell has electromotive force (e.m.f.) 1.50 V and internal resistance 1.2 Ω. It is connected in a circuit as shown below.
  • What is the terminal potential difference (p.d.) of the cell in this circuit? (1)
  • A. 0.15 V
  • B. 1.35 V
  • C. 1.50 V
  • D. 1.65 V
  • Result: B
  • Explanation:
  • Cell with internal resistance:
  • Given Data:
  • [math]E=1.50\text{V}[/math]
  • [math]r=1.2\Omega[/math]
  • (From the circuit, the current is 0.125 A.)
  • Terminal P.d:
  • [math]V=E-Ir[/math]
  • [math]V=1.50-(0.125)(1.2)[/math]
  • [math]V=1.35\text{V}[/math]
  • Terminal potential difference equals emf minus the lost volts inside the cell.
  • 8. A student uses a tube closed at one end to determine the speed of sound. A stationary wave is formed within the tube.
  • Which of the following quantities can be used to determine the speed of sound in air? (1)
  • A. length of air column in tube, number of antinodes
  • B. length of air column in tube, fundamental frequency
  • C. number of nodes, number of antinodes
  • D. number of nodes, resonant frequency.
  • Result: B
  • Explanation:
  • For a tube closed at one end:
  • [math]L=\frac{\lambda}{4}[/math]
  • Thus,
  • [math]v=f\lambda[/math]
  • Knowing the air-column length and resonant frequency allows the wavelength and speed to be calculated.
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