When lead, 81Pb214, emits a ß- particle, the resultant nucleus will be:
Correct answer: C. 82Pb214
- A. 83Bi214
- B. 84Po214
- C. 82Pb214
- D. 41TI214
Explanation
In beta decay, a neutron in the nucleus is transformed into a proton and a beta particle (an electron) is emitted. This results in an increase in the atomic number by 1 while the mass number remains unchanged. Thus, when lead, 81Pb214, emits a beta particle, it transforms into 82Pb214. The other options are incorrect as they either show an incorrect change in atomic number or involve elements that do not result from beta decay of lead.
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About Radioactive Decay
Radioactive decay is the spontaneous transformation of unstable nuclei, described by decay constant, activity, half life and the exponential decay law. Work includes alpha, beta and gamma emissions, decay equations and remaining nuclei, with half life distinguished from the time required for complete decay.
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