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Half-life for radioactive 14C is 5760 years. In how many years, 200 mg of 14C will be reduced to 25 mg?

1. 5760 years
2. 11520 years
3. 17280 years
4. 23040 years

Correct answer: 11520 years

Solution

The half-life formula is used: N = N0 * (1/2)^(t/t1/2). Here, N0 = 200 mg, N = 25 mg, and t1/2 = 5760 years. Solving for t, we find t = 11520 years.

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