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Half life for Carbon-14 is 5,730 years. If yous tart out with 50 grams, how long will...
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The half life of carbon 14 is 5730 years. If the starting amount of C-14 is P, the amount after t years is given by `A = P*(1/2)^(t/5730)`
Let the time taken for 50 g to decay to 2 g be t:
2 = `50*(1/2)^(t/5730)`
=> `log(1/25) = (t/5730)*log(1/2)`
=> `t = 5730*(log(1/25))/(log(1/2))`
=> t `~~` 26609 years
The time in which 50 g of C-14 decays to 2 g is approximately 26609 years.
Posted by justaguide on August 27, 2013 at 6:49 PM (Answer #1)
High School Teacher
To solve for the number of years it takes to the carbon to decay from 50g to 2g, plug-in A=2 and P=50 to the given formula.
Then, divide both sides by 50 to have (1/2)^(t/5730) at the right side of the equation.
To remove the t in the exponent, apply the loagrithm property ln `a^m = mlna` . So, take the natural logarithm of both sides of the equation.
Then, isolate t. To do so, multiply both sides by 5730.
`5730*ln(1/25) = t/5730 ln(1/2) * 5730`
And, divide both sides by ln(1/2).
Rounding off to nearest whole number, it becomes:
Hence, it takes 22,609 years for the Carbon-14 to decay from 50g to 2g.
Posted by mjripalda on August 28, 2013 at 1:33 AM (Answer #2)
First, we sub in 50 for P and 2 for A. Then, we solve for t.
First, divide each side by 50. So:
2/50 = (1/2)^t/5730
Then, take the log (base 10) of each side:
log(2/50) = log((1/2)^t/5730)
With this, though, the exponent on the right comes down in front of the log, as a factor, a multiplier. So, we have:
log(2/50) = (t/5730)*log(1/2)
Then, we can divide each side by log(1/2)"
[log(2/50)]/[log(1/2)] = t/5730
The right side is 4.644. Then, multiply each side by 5730, giving t = 26,610.12. So, it would take 26,610.12 years to get 2 grams.
I hope this helps, Kristen. Good luck.
Posted by steveschoen on August 28, 2013 at 12:56 AM (Answer #1)
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