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Q.

84Po218t1/2=183sec decay to Pb82t1/2=161sec by α-emission, while Pb214 is a β-emitter. In an experiment starting with 1 mole of pure Po218, how much time would be required for the number of nuclei of 82Pb214 to reach maximum?

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a

147.5

b

247.5

c

182

d

304

answer is B.

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Detailed Solution

84Po218⟶82Pb214⟶83Bi214+−1e0Where K1=0.693183=0.0038;K2=0.693161=0.0043Pb214 to reach max. no. of nucleitmax=1K1−K2ln⁡K1K2=247.5sec

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84Po218t1/2=183sec decay to Pb82t1/2=161sec by α-emission, while Pb214 is a β-emitter. In an experiment starting with 1 mole of pure Po218, how much time would be required for the number of nuclei of 82Pb214 to reach maximum?