Nuclear Stability: Quiz
Seven questions on the band of stability, N/Z ratios, predicting decay modes and decay series, with explanations.
Nuclear Stability: Quiz
7 questions
The strong force attracts all nucleons over very short distances, overcoming the repulsion between protons.
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Answer: The strong nuclear force
The strong force attracts all nucleons over very short distances, overcoming the repulsion between protons.
Too many neutrons: a neutron changes into a proton (β⁻), lowering N/Z towards the band of stability.
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Answer: Beta-minus decay
Too many neutrons: a neutron changes into a proton (β⁻), lowering N/Z towards the band of stability.
Too many protons for its neutrons: a proton becomes a neutron by emitting a positron. This is why F-18 is used in PET scans.
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Answer: Positron emission
Too many protons for its neutrons: a proton becomes a neutron by emitting a positron. This is why F-18 is used in PET scans.
Polonium (Z = 84) is beyond Z = 83, so it is too large and decays by alpha emission. C-14 and Co-60 are β⁻ emitters; Na-22 is a β⁺ emitter.
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Answer: Polonium-210
Polonium (Z = 84) is beyond Z = 83, so it is too large and decays by alpha emission. C-14 and Co-60 are β⁻ emitters; Na-22 is a β⁺ emitter.
N = 206 − 82 = 124; N/Z = 124 ÷ 82 = 1.51. Heavy stable nuclei need about 1.5 neutrons per proton.
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Answer: 1.51
N = 206 − 82 = 124; N/Z = 124 ÷ 82 = 1.51. Heavy stable nuclei need about 1.5 neutrons per proton.
Only alpha decay changes the mass number: (232 − 208) ÷ 4 = 6 alpha decays.
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Answer: 6
Only alpha decay changes the mass number: (232 − 208) ÷ 4 = 6 alpha decays.
6 alpha decays lower Z by 12, to 78. Lead has Z = 82, so Z must rise by 4: 4 β⁻ decays.
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Answer: 4
6 alpha decays lower Z by 12, to 78. Lead has Z = 82, so Z must rise by 4: 4 β⁻ decays.