The radioactivity of a certain radioactive el | vedclass

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(C) The law of radioactive decay is given by $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $T$ is the half-life.Given that the radioactivity dro

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$5$

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(C) The law of radioactive decay is given by $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $T$ is the half-life.Given that the radioactivity drops to $1/64$ of its initial value in $t = 30 \, s$,we have $\frac{N}{N_0} = \frac{1}{64}$.Substituting this into the formula: $\frac{1}{64} = \left( \frac{1}{2} \right)^{30/T}$.Since $64 = 2^6$,we can write $\left( \frac{1}{2} \right)^6 = \left( \frac{1}{2} \right)^{30/T}$.Equating the exponents: $6 = \frac{30}{T}$.Therefore,$T = \frac{30}{6} = 5 \, s$.

The radioactivity of a certain radioactive element drops to $1/64$ of its initial value in $30 \, s$. Its half-life is ......... $s$.

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