1 , mg of gold undergoes decay with a 2.7 , t | vedclass

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(D) The formula for radioactive decay is given by $N = N_0 \left( \frac{1}{2} \right)^n$, where $n = \frac{t}{T_{1/2}}$.Given:Initial amount $N_0 = 1

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$0.125$

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(D) The formula for radioactive decay is given by $N = N_0 \left( \frac{1}{2} ight)^n$, where $n = \frac{t}{T_{1/2}}$.Given:Initial amount $N_0 = 1 \, mg$Total time $t = 8.1 \, 	ext{days}$Half-life $T_{1/2} = 2.7 \, 	ext{days}$First, calculate the number of half-lives $n$:$n = \frac{8.1}{2.7} = 3$Now, substitute the values into the decay formula:$N = 1 	imes \left( \frac{1}{2} ight)^3$$N = \frac{1}{8} = 0.125 \, mg$Thus, the amount left after $8.1 \, 	ext{days}$ is $0.125 \, mg$.

$1 \, mg$ of gold undergoes decay with a $2.7 \, \text{days}$ half-life period. The amount left after $8.1 \, \text{days}$ is ......... $mg$.

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