Radioactivity

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. 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.

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

1. Eight days is four half-lives for A; 80 years is four for B. Each activity becomes 800/2⁴ = 50 Bq, despite very different calendar times.
2. A transmission factor 1/32 is (1/2)⁵, so five 1 cm layers are needed in the simplified shielding model.
3. Decay storage and shielding reduce different aspects of exposure: elapsed time reduces activity, while shielding reduces radiation reaching a location without changing the decay rate.
4. A becquerel is one decay per second, not a unit of absorbed or biologically weighted dose. Decay energy, radiation type and uptake can differ.
5. A long-lived sample may need containment for far longer, while a short-lived sample can have intense early emission. Daughter products can complicate both histories.
6. Compare measured dose rates under specified geometry, containment integrity, transport and storage requirements, and credible contamination pathways. The simple half-value model alone cannot certify a storage design.

Answer (a) Both specified intervals give 50 Bq; five half-value layers require 5 cm. (b) Equal Bq does not imply equal dose or risk; assess radiation, geometry, duration and pathways as well as activity.

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