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]