Nuclear and quantum 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. Most projectiles cross the foil with little change in direction, so atoms are not solid dense obstacles across their full dimensions.
2. A rare close encounter with concentrated positive nuclear charge can strongly repel an alpha particle. Very large deflections require a strong local interaction and a sufficiently massive scattering centre.
3. A diffuse positive-charge model would spread the repulsion over the atomic volume and struggles to explain the observed rare strong deflections. The evidence supports a compact nucleus, while this count table alone does not determine an exact nuclear radius.
4. Atomic number gives 79 protons. Neutrons are A−Z = 197−79 = 118; neutrality requires 79 electrons.
5. A +3e ion has lost three electrons and has 76; its nucleus is unchanged. An isotope has the same proton number but a different neutron number.
6. Alpha emission removes A = 4 and Z = 2: ²⁰¹₈₁X → ¹⁹⁷₇₉Au + ⁴₂He. Both nucleon and charge numbers balance.

Answer (a) Most atomic volume is sparsely occupied; positive charge and most mass are concentrated in a small nucleus. (b) Neutral Au-197: 79 p, 118 n, 79 e; Au³⁺ has 76 e. Alpha daughter: A = 197, Z = 79.

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

1. Use λ = c/f: the lower frequency has wavelength 6.0×10⁻⁷ m and the higher 3.0×10⁻⁷ m.
2. Ephoton = hf gives 3.313×10⁻¹⁹ and 6.626×10⁻¹⁹ J. Dividing equal total pulse energy by these values gives 2.0×10¹³ and 1.0×10¹³ photons.
3. 600 nm is visible light and 300 nm is ultraviolet. Both propagate at c in vacuum; the higher-frequency pulse contains fewer photons for the same total energy.
4. Spatial interference patterns follow phase and superposition, which a simple stream of classical particles does not explain.
5. Localized detector events with frequency-dependent energy are described by photons. These are complementary descriptions of quantum light rather than literal tiny balls riding classical waves.
6. Absorption, penetration, dose and photon energy depend on wavelength and material. Ultraviolet can cause photochemical damage; exposure assessment cannot be inferred from vacuum speed alone.

Answer (a) 600 nm, 3.313×10⁻¹⁹ J, 2.0×10¹³ photons; 300 nm, 6.626×10⁻¹⁹ J, 1.0×10¹³ photons. (b) Different models describe different observations; common vacuum speed does not imply identical interaction or hazard.

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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.