The ratio of the abundance of carbon-14 to carbon-12 in asample of dead wood is

Question

The ratio of the abundance of carbon-14 to carbon-12 in asample of dead wood is one quarter the ratio for living wood. Ifthe half-life of carbon-14 is 5730 years, which one of thefollowing expressions determines how many years ago the wooddied?

a. 0.75 × 5730
b. 2 × 5730
c. 0.50 × 5730
d. 0.25 × 5730
e. 4 × 5730 The ratio of the abundance of carbon-14 to carbon-12 in asample of dead wood is one quarter the ratio for living wood. Ifthe half-life of carbon-14 is 5730 years, which one of thefollowing expressions determines how many years ago the wooddied? 0.75 × 5730 2 × 5730 0.50 × 5730 0.25 × 5730 4 × 5730

Explanation / Answer

We know that N = No * e-t -----------(1) here,N is the number of undecayed radioactive nuclie presentat some instant,No represents the number of undecayedradioactive nuclie at t = 0. solving equation (1) for t we get t = (1/) * ln(No/N) ------------(2) the disintegration constant s given by = (0.693/T1/2) ------------(3) T1/2 = 5730 from equations (2) and (3) we get t = (1/(0.693/T1/2)) * ln(No/N) or t = (T1/2/0.693) * ln(No/N) here,N = (No/4) or t = T1/2 * [ln(No/N)/0.693] =T1/2 * [ln(No/(No/4))/0.693] =T1/2 * [ln(4)/0.693] = 2 * T1/2 T1/2 = 5730 years or t = 2 * 5730 here,N is the number of undecayed radioactive nuclie presentat some instant,No represents the number of undecayedradioactive nuclie at t = 0. solving equation (1) for t we get t = (1/) * ln(No/N) ------------(2) the disintegration constant s given by = (0.693/T1/2) ------------(3) T1/2 = 5730 from equations (2) and (3) we get t = (1/(0.693/T1/2)) * ln(No/N) or t = (T1/2/0.693) * ln(No/N) here,N = (No/4) or t = T1/2 * [ln(No/N)/0.693] =T1/2 * [ln(No/(No/4))/0.693] =T1/2 * [ln(4)/0.693] = 2 * T1/2 T1/2 = 5730 years or t = 2 * 5730

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