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(D) The law of radioactive decay is given by $\frac{N}{N_0} = (\frac{1}{2})^{t/T}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ i

Vedclass · Accepted Answer

$192$

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(D) The law of radioactive decay is given by $\frac{N}{N_0} = (\frac{1}{2})^{t/T}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T$ is the half-life.Given,$\frac{N}{N_0} = \frac{1}{16}$ and $T = 48 	ext{ hours}$.Substituting the values: $\frac{1}{16} = (\frac{1}{2})^{t/48}$.Since $\frac{1}{16} = (\frac{1}{2})^4$,we have $(\frac{1}{2})^4 = (\frac{1}{2})^{t/48}$.Equating the exponents: $4 = \frac{t}{48}$.Therefore,$t = 4 	imes 48 = 192 	ext{ hours}$.

The half-life of a radioactive substance is $48$ hours. How much time will it take to disintegrate to its $\frac{1}{16}$th part?

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