The half-life of c-14 is 5470 years. if a particular archaeological sample has one-quarter of its original radioactivity remaining, what is the best estimate for its age? 5470 years 10940 years 16410 years 21880 years

When there remains one-quarter of the sample, the age of the sample is 10940 years

Explanation:

Step 1: Data given

The half- life time is the time required for a quantity to reduce to half of its initial value.

The half-life of C-14 is 5470 years.

This means after 5470 years there remains half of the C-14 sample.

When there will pass another half-life cyclus, half of the sample will remain.

Half of 50 % = 25% = one- quarter

This means 2 half-lives should have passed to remain a quarter of the sample.

Step 2: Calculate it's age

t/(t/1/2) = 2

⇒ with t = the age (or time) of the sample

⇒ with t(1/2) = the half-life time of the sample = 5470 years

⇒ with 2 = the number of half- lives passed  to remain one quarter of sample

t/5470 = 2

t = 2*5470 = 10940 years

When there remains one-quarter of the sample, the age of the sample is 10940 years

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years