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(C) The radioactive decay follows the formula $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$,where $N_0$ is the initial mass,$t$ is the tim

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$\frac{125}{4} 	ext{ mg}$

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(C) The radioactive decay follows the formula $N(t) = N_0 \left(\frac{1}{2}ight)^{\frac{t}{T_{1/2}}}$,where $N_0$ is the initial mass,$t$ is the time elapsed,and $T_{1/2}$ is the half-life period.Given: $N_0 = 1000 	ext{ mg}$,$t = 50 	ext{ days}$,and $T_{1/2} = 10 	ext{ days}$.The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{50}{10} = 5$.The remaining mass $N(t) = 1000 	imes \left(\frac{1}{2}ight)^5$.$N(t) = 1000 	imes \frac{1}{32} = \frac{1000}{32} 	ext{ mg}$.Dividing both numerator and denominator by $8$,we get $N(t) = \frac{125}{4} 	ext{ mg}$.

$A$ radioactive sample has a half-life period of $10 \text{ days}$. If a sample originally has a mass of $1000 \text{ mg}$,then the mass remaining after $50 \text{ days}$ is:

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