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(C) The radioactive decay law 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 amou

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$\frac{1}{2}\, hour$

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(C) The radioactive decay law 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: $N_0 = 16\, g$,$N = 1\, g$,and $t = 2\, hours$.Substituting these values into the formula:$1 = 16 \left( \frac{1}{2} \right)^{2/T_{1/2}}$$\frac{1}{16} = \left( \frac{1}{2} \right)^{2/T_{1/2}}$$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{2/T_{1/2}}$Equating the exponents:$4 = \frac{2}{T_{1/2}}$$T_{1/2} = \frac{2}{4} = \frac{1}{2}\, hour$.

$A$ $16\, g$ sample of a radioactive element is taken from Bombay to Delhi in $2\, hours$ and it is found that $1\, g$ of the element remains (undisintegrated). The half-life of the element is:

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