The half-life of a radioactive element is 10 | vedclass

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(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.Given, $T_{1/2} = 10 \, days$ and $\frac{N}{N_0} =

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$33$

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(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} ight)^{\frac{t}{T_{1/2}}}$.Given, $T_{1/2} = 10 \, days$ and $\frac{N}{N_0} = \frac{1}{10}$.Substituting these values: $\frac{1}{10} = \left( \frac{1}{2} ight)^{\frac{t}{10}}$.Taking the natural logarithm on both sides: $\ln(10^{-1}) = \frac{t}{10} \ln(2^{-1})$.$- \ln(10) = - \frac{t}{10} \ln(2)$.$t = 10 	imes \frac{\ln(10)}{\ln(2)} = 10 	imes \frac{2.303}{0.693} \approx 10 	imes 3.32$.$t \approx 33.2 \, days$.Therefore, the time required is approximately $33 \, days$.

The half-life of a radioactive element is $10 \, days$. The time required for the mass of the sample to reduce to $\frac{1}{10}$ of its initial mass is approximately ........ days.

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