The half-life of the isotope _{11}Na^{24} is | vedclass

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(D) The fraction of the isotope that has decayed is $\frac{7}{8}$.Therefore,the fraction of the isotope that remains undecayed is $N/N_0 = 1 - \frac{7

Vedclass · Accepted Answer

$45$

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(D) The fraction of the isotope that has decayed is $\frac{7}{8}$.Therefore,the fraction of the isotope that remains undecayed is $N/N_0 = 1 - \frac{7}{8} = \frac{1}{8}$.Using the radioactive decay law: $\frac{N}{N_0} = (\frac{1}{2})^{t/T}$,where $T = 15 \, hrs$ is the half-life.Substituting the values: $\frac{1}{8} = (\frac{1}{2})^{t/15}$.Since $\frac{1}{8} = (\frac{1}{2})^3$,we have $(\frac{1}{2})^3 = (\frac{1}{2})^{t/15}$.Equating the exponents: $3 = \frac{t}{15}$.Solving for $t$: $t = 3 	imes 15 = 45 \, hrs$.

The half-life of the isotope $_{11}Na^{24}$ is $15 \, hrs$. How much time does it take for $\frac{7}{8}$ of a sample of this isotope to decay?

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