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?

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

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?

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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$_{84} {Po}^{218} (t_{1 / 2} = 183 \sec) decay to {Pb}^{82} (t_{1 / 2} = 161 \sec) by$ $α$ -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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$_{84} {Po}^{218} ⟶_{82} {Pb}^{214} ⟶_{83} {Bi}^{214} +_{- 1} e^{0}$

Where $K_{1} = \frac{0 .693}{183} = 0 .0038; K_{2} = \frac{0 .693}{161} = 0 .0043$

Pb²¹⁴ to reach max. no. of nuclei

$t_{max} = \frac{1}{K_{1} - K_{2}} \ln \frac{K_{1}}{K_{2}} = 247 .5 \sec$

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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?

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$_{84} {Po}^{218} (t_{1 / 2} = 183 \sec) decay to {Pb}^{82} (t_{1 / 2} = 161 \sec) by$ $α$ -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?