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Answer :
Final answer:
The activity of 2 mg of Au-198, with a half-life of 3 days, is 2.52x10^-34 g.
Explanation:
The activity of a radioactive substance is the rate at which it undergoes decay. In this case, we are given that the half-life of Au-198 is 3 days. The half-life is the amount of time it takes for half of the radioactive substance to decay. So, after each half-life, the amount of Au-198 remaining is halved.
To determine the activity of 2 mg (0.002 g) of Au-198, we can use the formula:
Activity = Initial Mass × (1/2)^(Number of Half-lives)
Since the atomic weight of Au-198 is 198 grams per mole, we can convert 2 mg to grams by dividing by 1000:
Initial Mass = 0.002 g
Now, we need to find the number of half-lives. Since the half-life is 3 days, we divide the total time by the half-life:
Number of Half-lives = Total Time / Half-life
= 0.002 g / 198 g/mol × 6.02x10^23 atoms/mol × 1 s / (3 days × 24 hours/days × 60 minutes/hour × 60 seconds/minute)
Calculating the values gives us:
Number of Half-lives = 0.002 g / (198 g/mol × 6.02x10^23 atoms/mol × 1 s / (3 days × 24 hours/days × 60 minutes/hour × 60 seconds/minute))
Number of Half-lives = 1.04x10^17 half-lives
Now, we can substitute the values into the activity formula:
Activity = 0.002 g × (1/2)^(1.04x10^17)
Activity = 0.002 g × 1.26x10^-31
Activity = 2.52x10^-34 g
Therefore, the activity of 2 mg of Au-198 is 2.52x10^-34 g.
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