Physics

1. 1.

   Parallel plates have a potential difference of 600 V and separation 0.020 m. Assume a uniform field. Calculate its magnitude and the magnitude of the force on a charge of +2.0 microcoulombs.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The field is the potential difference per metre, not voltage multiplied by distance.
   2. Convert 2.0 microcoulombs to 2.0 x 10^-6 C before using F = qE.
   3. A positive charge experiences force along the electric field, from the positive plate towards the negative plate.

   Marking points

   • E = V/d = 600/0.020.
   • E = 3.0 x 10^4 V m^-1.
   • F = qE = 0.060 N.

   Examiner tip: Convert both the plate separation and the charge to SI units first.

2. 2.

   A detector reads 820 counts per minute initially and 120 counts per minute 6.0 hours later. Background is constant at 20 counts per minute. Assuming exponential decay, find the sample's half-life.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Subtract the same background from both readings; the background is not part of the decaying sample.
   2. Compare corrected rates: 800 -> 400 -> 200 -> 100. This is three successive halvings.
   3. Divide the total elapsed time by three. Count rates are proportional to activity under unchanged measurement conditions.

   Marking points

   • Corrected count rates are 800 and 100 counts per minute.
   • The sample rate has fallen to 1/8 of its initial value.
   • 1/8 = (1/2)^3, so three half-lives have elapsed.
   • Half-life = 6.0/3 = 2.0 hours.

   Examiner tip: Taking 120/820 without removing background gives an incorrect decay ratio.

3. 3.

   A metal has work function 2.0 eV. Light of wavelength 400 nm falls on it. Use h = 6.63 x 10^-34 J s, c = 3.00 x 10^8 m s^-1 and 1 eV = 1.60 x 10^-19 J. Find maximum photoelectron kinetic energy in eV and stopping potential. Explain whether doubling intensity changes stopping potential.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Convert wavelength to 4.00 x 10^-7 m. hc/lambda = 4.9725 x 10^-19 J, or 3.1078 eV.
   2. One photon supplies one electron. Subtract the work function: Kmax = 3.1078 - 2.0, about 1.1 eV at the given precision.
   3. Stopping potential removes that energy through eV. Intensity changes photon arrival rate, not each photon's energy.

   Marking points

   • Photon energy hc/lambda is approximately 3.11 eV.
   • Maximum kinetic energy is approximately 1.1 eV.
   • Stopping potential is approximately 1.1 V.
   • Doubling intensity at fixed frequency does not change the maximum energy or stopping potential; it can increase photocurrent.

   Examiner tip: Greater intensity does not mean greater photon energy. Photon energy depends on frequency.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.