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(B) The law of radioactive decay is given by the formula: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T}$, where $N$ is the remaining amount, $N_0$

Vedclass · Accepted Answer

$30 \, minutes$

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(B) The law of radioactive decay is given by the formula: $\frac{N}{N_0} = \left( \frac{1}{2} ight)^{t/T}$, where $N$ is the remaining amount, $N_0$ is the initial amount, $t$ is the elapsed time, and $T$ is the half-life.Given that $\frac{N}{N_0} = \frac{1}{16}$ and $t = 2 \, hours$.Substituting these values into the formula:$\frac{1}{16} = \left( \frac{1}{2} ight)^{2/T}$Since $\frac{1}{16} = \left( \frac{1}{2} ight)^4$, we have:$\left( \frac{1}{2} ight)^4 = \left( \frac{1}{2} ight)^{2/T}$Equating the exponents:$4 = \frac{2}{T}$$T = \frac{2}{4} = 0.5 \, hours$.Converting to minutes: $0.5 	imes 60 = 30 \, minutes$.

After two hours, one-sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half-life of the isotope is:

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