Modern Physics

1. Energy drops are (−1.5)−(−3.0)=1.5 eV, (−3.0)−(−6.0)=3.0 eV and (−1.5)−(−6.0)=4.5 eV.
2. A direct transition emits one 4.5 eV photon; a two-step cascade emits separate 1.5 and 3.0 eV photons, not one of each plus 4.5 eV in the same decay.
3. The greatest photon energy gives shortest wavelength: λ = hc/E = (4.14×10⁻¹⁵)(3.0×10⁸)/4.5≈2.76×10⁻⁷ m.
4. A bound–bound absorption must match an allowed energy difference; the listed ground-state excitation gaps are 3.0 and 4.5 eV, not 2.0 eV.
5. Ground-state ionisation needs 6.0 eV. Absorbing 6.5 eV can free the electron with 0.5 eV excess kinetic energy, neglecting recoil.
6. Discrete lines support quantised energy changes. They do not show electrons following literal tiny planetary trajectories, and a real atom may have selection rules absent from this idealised model.

Answer (a) Possible photons: 1.5, 3.0 and 4.5 eV; shortest wavelength≈276 nm. (b) A 2.0 eV photon matches no listed excitation; a 6.5 eV photon can ionise the atom, leaving 0.5 eV kinetic energy.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Einstein's relation gives Vs = (h/e)f−φ/e. Graph slope is 0.414/(1.0×10¹⁴) = 4.14×10⁻¹⁵ V s.
2. Thus h = e×slope = 6.624×10⁻³⁴ J s. At threshold, φ = hf0 = 3.312×10⁻¹⁹ J, or 2.07 eV.
3. At 8.0×10¹⁴ Hz, Kmax = eVs = (1.60×10⁻¹⁹)(1.242)≈1.987×10⁻¹⁹ J. The graph intercept is an energy-per-charge condition, not a delay.
4. At fixed frequency, each photon carries the same hf. Doubling intensity supplies more photons per second, which can increase photocurrent if collection and material conditions remain unchanged.
5. Stopping potential depends on the maximum energy per emitted electron, so it does not double with intensity.
6. Below threshold each photon lacks the work-function energy. Increasing photon arrival rate does not fix that within the stated single-photon model; multiphoton regimes are outside this approximation.

Answer (a) h≈6.624×10⁻³⁴ J s; work function≈3.312×10⁻¹⁹ J = 2.07 eV; Kmax≈1.987×10⁻¹⁹ J = 1.242 eV. (b) Above threshold, emission rate can rise without changing Kmax; below threshold the ideal one-photon model predicts no photoemission.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Subtract 40 from each raw reading before analysing source decay. The corrected signal halves every 5 min.
2. At 30 min six half-lives have elapsed, giving expected net counts 320/2⁶ = 5.
3. The detector still records background, so expected raw counts are 5+40 = 45. A real individual reading will fluctuate randomly about this expectation.
4. Activity describes decays per second, while counts depend on detector efficiency, radiation type, shielding and geometry. The source can remain active when this setup cannot clearly distinguish it from background.
5. Longer matched source and background counting periods reduce relative random uncertainty, while repeated readings reveal scatter. Keep distance and shielding fixed for a fair decay comparison.
6. Risk also depends on exposure time, energy, biological uptake and contamination pathways. Use the relevant handling controls rather than inferring safety from one low count.

Answer (a) Net counts are 320, 160, 80, 40 and 20; half-life = 5 min; expected raw count at 30 min = 45. (b) Near-background readings limit detectability; they do not by themselves establish absence of activity or a safe exposure.

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1. Nucleon number gives 235+1 = 141+92+x, hence x = 3. Charge checks as 92 = 56+36; neutrons carry no charge.
2. The measured population ratio is 110/100 = 121/110 = 1.10, so this illustrative sequence grows by 10% per generation.
3. A moderator slows neutrons in a thermal reactor without being intended to absorb many; control rods absorb neutrons to adjust the effective multiplication. These functions are not interchangeable.
4. Energy per fission is 200×10⁶×1.60×10⁻¹⁹ = 3.20×10⁻¹¹ J.
5. Rate = power/energy per event = 1.0×10⁹/(3.20×10⁻¹¹) = 3.125×10¹⁹ fissions s⁻¹.
6. The products have lower total rest mass and greater total binding energy; the mass difference appears as released energy. Nucleon and charge conservation do not imply equality of the sum of the separate rest masses.

Answer (a) x = 3; effective multiplication factor = 1.10; control must bring it toward 1 for steady operation. (b) Fission rate = 3.125×10¹⁹ s⁻¹.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Convert 26.7 MeV to joules: E = 26.7(1.60×10⁻¹³) = 4.272×10⁻¹² J.
2. The decrease in total rest mass is Δm = E/c² = 4.272×10⁻¹²/(9.0×10¹⁶)≈4.75×10⁻²⁹ kg per net cycle.
3. Cycle rate is L/E = 4.0×10²⁶/(4.272×10⁻¹²)≈9.36×10³⁷ s⁻¹. Nuclear binding-energy differences, rather than chemical combustion, power this process.
4. Each cycle uses about 6.68×10⁻²⁷ kg of hydrogen, so consumption is (9.36×10³⁷)(6.68×10⁻²⁷)≈6.25×10¹¹ kg s⁻¹.
5. Time is available fuel divided by its consumption rate: 2.0×10²⁹/(6.25×10¹¹)≈3.20×10¹⁷ s≈1.01×10¹⁰ yr.
6. Luminosity changes as the star evolves, only part of its hydrogen is accessible to core fusion, and some fusion energy can escape in neutrinos. A full stellar model must account for structure, composition and evolving reactions.

Answer (a) Energy = 4.272×10⁻¹² J; mass defect≈4.75×10⁻²⁹ kg; rate≈9.36×10³⁷ cycles s⁻¹. (b) Fuel consumption≈6.25×10¹¹ kg s⁻¹; idealised lifetime≈1.01×10¹⁰ years.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.