In how many years will a 80 mg of sample of C14 with t1/2 5760 years be reduced to 20 mg
See your post above. Same kind of problem. Find k, then plug in the numbers for the second part of the problem.
To calculate the number of years it takes for a sample of C14 to be reduced to a specific amount, we can use the formula for exponential decay:
N(t) = N(0) * (1/2)^(t / t1/2)
Where:
N(t) = remaining amount of the sample at time t
N(0) = initial amount of the sample
t = time elapsed
t1/2 = half-life of the sample
Given that the initial amount N(0) is 80 mg, the remaining amount N(t) is 20 mg, and the half-life t1/2 is 5760 years, we can rearrange the formula to solve for t:
20 mg = 80 mg * (1/2)^(t / 5760 years)
Dividing both sides of the equation by 80 mg, we get:
(1/2)^(t / 5760 years) = 1/4
To isolate the exponential term, we take the logarithm of both sides. Since logarithms with base 2 will simplify the equation due to the power of 2, we can use the logarithm base 2:
log2((1/2)^(t / 5760 years)) = log2(1/4)
Using logarithm properties, we can bring down the exponent:
(t / 5760 years) * log2(1/2) = log2(1/4)
The logarithm base 2 of 1/2 equals -1:
(t / 5760 years) * (-1) = log2(1/4)
Simplifying further:
t / 5760 years = log2(1/4) * (-1)
Since log2(1/4) = -2:
t / 5760 years = -2
To solve for t, we multiply both sides by 5760 years:
t = -2 * 5760 years
However, we have a negative value here, which doesn't make sense in this context. Decay time cannot have a negative value.
Thus, the equation cannot be solved, and we cannot determine the number of years it will take for the C14 sample to be reduced to 20 mg in this case.