The half-life of polonium is 140 , days. Afte | vedclass

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(C) The law of radioactive decay is given by the formula: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{1/2}}$,where $N$ is the remaining amount,$

Vedclass · Accepted Answer

$560$

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(C) The law of radioactive decay is given by the formula: $\frac{N}{N_0} = \left( \frac{1}{2} ight)^{t/T_{1/2}}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.Given: $N_0 = 16 \, g$,$N = 1 \, g$,and $T_{1/2} = 140 \, days$.Substituting the values: $\frac{1}{16} = \left( \frac{1}{2} ight)^{t/140}$.Since $\frac{1}{16} = \left( \frac{1}{2} ight)^4$,we have $\left( \frac{1}{2} ight)^4 = \left( \frac{1}{2} ight)^{t/140}$.Equating the exponents: $4 = \frac{t}{140}$.Therefore,$t = 4 	imes 140 = 560 \, days$.

The half-life of polonium is $140 \, days$. After how many days will $16 \, g$ of polonium be reduced to $1 \, g$ (or $15 \, g$ will decay)?

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