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The half-life of a radioactive element is $10 \mathrm{~h}$. The fraction of initial radioactivity of the element that will remain after $40 \mathrm{~h}$ is
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The correct answer is:
$\frac{1}{16}$
After 4 half-lifes
$\frac{N_0}{2^4}=\frac{N_0}{16}$
$\frac{N_0}{2^4}=\frac{N_0}{16}$
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