After 40 , days,the 1/16th part of a radioact | vedclass

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

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$10$

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(C) The law of radioactive decay is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$,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 that the remaining amount is $1/16$th of the initial amount,we have $\frac{N}{N_0} = \frac{1}{16}$.Substituting the values into the formula: $\frac{1}{16} = \left( \frac{1}{2} \right)^{40/T_{1/2}}$.Since $\frac{1}{16} = \left( \frac{1}{2} \right)^4$,we can equate the exponents: $4 = \frac{40}{T_{1/2}}$.Solving for $T_{1/2}$,we get $T_{1/2} = \frac{40}{4} = 10 \, days$.

After $40 \, days$,the $1/16$th part of a radioactive element remains undecayed. What is its half-life (in $, days$)?

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