The half-life of Bi^{210} is 5 days. If we st | vedclass

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(C) The formula for the number of radioactive atoms remaining after time $t$ is given by $N_t = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N_0$ is t

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$12,500$

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(C) The formula for the number of radioactive atoms remaining after time $t$ is given by $N_t = N_0 \left( \frac{1}{2} ight)^{t/T}$,where $N_0$ is the initial number of atoms,$t$ is the elapsed time,and $T$ is the half-life.Given:Initial number of atoms $N_0 = 50,000$Elapsed time $t = 10 	ext{ days}$Half-life $T = 5 	ext{ days}$Substituting these values into the formula:$N_t = 50,000 	imes \left( \frac{1}{2} ight)^{10/5}$$N_t = 50,000 	imes \left( \frac{1}{2} ight)^2$$N_t = 50,000 	imes \frac{1}{4}$$N_t = 12,500$Therefore,the number of atoms left after $10$ days is $12,500$.

The half-life of $Bi^{210}$ is $5$ days. If we start with $50,000$ atoms of this isotope,the number of atoms left over after $10$ days is:

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