Physics

1. 1.

   Find energy in joules of a photon of frequency 5.00e14 Hz. Use h = 6.63e-34 J s. Explain why doubling beam intensity at fixed frequency does not double photon energy.

   [3 marks]

   Answer explanation

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

   1. Multiply Planck's constant by frequency, retaining full precision until rounding.
   2. A brighter monochromatic beam supplies more photons, each with the same hf.

   Marking points

   • E = hf.
   • E = 3.32e-19 J (3 significant figures).
   • Intensity changes photon rate per unit area, not energy per photon at fixed frequency.

   Examiner tip: Photon energy is set by frequency, not brightness.

2. 2.

   A stationary uranium-238 nucleus emits an alpha particle. State the daughter's mass and proton numbers and explain why it recoils. Neglect photon emission.

   [3 marks] · no calculator

   Answer explanation

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

   1. Subtract alpha's nucleon and proton counts separately; the daughter is not another uranium isotope.
   2. Different masses mean equal momentum magnitudes do not imply equal recoil speeds.

   Marking points

   • Alpha has mass number 4 and proton number 2; uranium has proton number 92.
   • Daughter has mass number 234 and proton number 90 (thorium).
   • The daughter recoils with momentum equal and opposite to alpha momentum, conserving initial zero momentum.

   Examiner tip: Conserve both nucleon number and charge in the decay equation.

3. 3.

   An atom emits a photon when its electron moves from -1.50 eV to -3.50 eV. Find wavelength using h = 6.63e-34 J s, c = 3.00e8 m s^-1 and 1 eV = 1.60e-19 J. Explain why the atom loses energy.

   [3 marks]

   Answer explanation

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

   1. Use initial minus final level for positive emitted photon energy: -1.50 - (-3.50).
   2. Convert eV to joules before dividing hc; the negative level labels do not mean negative photon energy.

   Marking points

   • Photon energy is 2.00 eV = 3.20e-19 J.
   • lambda = hc/E = 622 nm (3 significant figures).
   • The final level is lower; energy difference leaves the atom as the photon.

   Examiner tip: Use the difference between levels, not the magnitude of either individual level.

4. 4.

   A sample initially contains 2.00e12 radioactive nuclei and has half-life 6.00 hours. Find decay constant in s^-1 and activity after 12.0 hours. Assume no daughter contribution. Give 3 significant figures.

   [3 marks]

   Answer explanation

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

   1. Activity counts decays per second, so convert the half-life to seconds before finding lambda.
   2. Use the remaining parent population and unrounded lambda; counting daughters would violate the stated model.

   Marking points

   • lambda = ln(2)/(6.00*3600) = 3.21e-5 s^-1.
   • After two half-lives N = 5.00e11.
   • Activity lambda N = 1.60e7 Bq (3 significant figures).

   Examiner tip: Becquerels require seconds, even when elapsed times are stated in hours.

5. 5.

   An electron initially at rest is accelerated through 150 V in vacuum. Find its de Broglie wavelength using e = 1.60e-19 C, m = 9.11e-31 kg, h = 6.63e-34 J s. Use non-relativistic mechanics and justify that assumption from its speed with c = 3.00e8 m s^-1.

   [3 marks]

   Answer explanation

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

   1. Electric potential energy supplies kinetic energy, and the classical momentum follows from KE = p^2/(2m).
   2. This is a matter-wave wavelength; hc/E is the photon formula and is not applicable to this electron.

   Marking points

   • KE = eV; momentum p = sqrt(2meV).
   • lambda = h/p = 1.00e-10 m (3 significant figures).
   • v = 7.26e6 m s^-1, about 0.0242c, so a non-relativistic approximation is reasonable.

   Examiner tip: Use the magnitude of electron charge for energy gained through the accelerating voltage.

6. 6.

   A nucleus with mass number 4 has mass defect 0.0300 u relative to separated nucleons. Use 1 u = 1.66e-27 kg, c = 3.00e8 m s^-1 and 1 MeV = 1.60e-13 J. Find total binding energy and binding energy per nucleon. Explain the sign of energy needed to separate it.

   [3 marks]

   Answer explanation

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

   1. Use nuclear masses consistently with the separated nucleons; the supplied defect already accounts for that comparison.
   2. The bound state has lower mass-energy. Separation reverses binding and therefore requires external energy, rather than releasing it.

   Marking points

   • Mass defect = 4.98e-29 kg; binding energy = delta m c^2.
   • Total binding energy = 4.48e-12 J = 28.0 MeV (3 significant figures).
   • Binding energy per nucleon = 7.00 MeV; positive energy must be supplied to separate the nucleus.

   Examiner tip: Divide total binding energy by mass number, not proton number.

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