The number of half-lives elapsed before 93.75 | vedclass

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(B) The number of half-lives $n$ can be calculated using the radioactive decay law: $N = N_0 \left( \frac{1}{2} \right)^n$ where $N$ is the amount rem

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The number of half-lives $n$ can be calculated using the radioactive decay law: $N = N_0 \left( \frac{1}{2} ight)^n$ where $N$ is the amount remaining after $n$ half-lives. Given that $93.75 \%$ of the sample has decayed,the amount remaining is: $N = (100 - 93.75) \% 	ext{ of } N_0 = 6.25 \% 	ext{ of } N_0 = \frac{6.25}{100} N_0 = \frac{1}{16} N_0$ Substituting this into the decay equation: $\frac{1}{16} N_0 = N_0 \left( \frac{1}{2} ight)^n$ $\frac{1}{16} = \left( \frac{1}{2} ight)^n$ Since $16 = 2^4$,we have: $\left( \frac{1}{2} ight)^4 = \left( \frac{1}{2} ight)^n$ Therefore,$n = 4$.

The number of half-lives elapsed before $93.75 \%$ of a radioactive sample has decayed is

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