Question

The half-life for the radioactive decay of c-14 is 5730 years and is independent of the initial concentration. how long does it take for 25% of the c-14 atoms in a sample of c-14 to decay? if a sample of c-14 initially contains 1.5 mmol of c-14, how many millimoles are left after 2255 years?

Answer 1

1. If 25% of the c-14 atoms in the sample decay, that means the final weight would be 100-25%= 75% of the initial weight. Then, the amount of time elapsed would be:

final weight= initial weight * 1/2^ (t/t1/2)
0.75* of the initial weight= initial weight * 2^-t/5730 ---->1/2 = 2^-1
0.75 = 2^-t/5730
log2 0.75 = log2 (2^-t/5730)
-0.415= -t/5730 ---->2^-0.415= 0.75
t=0.415*5730
t=2378 years

2. if a sample of c-14 initially contains 1.5 mmol of c-14, how many millimoles are left after 2255 years?


final weight= initial weight * 1/2^ (t/t1/2)
final weight= 1.5mol * 1/2^ (2255/5730)
final weight= 1.5mol* 1/2^0.39354
final weight= 1.5mol* 0.761= 1.14 mol

Answer 2

(1) The time passed by the sample is
`2.4* 10^3\text{ years}`

(2) The amount left after decay process is 1.14 mmol.

Explanation :

Part 1 :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

`k=(0.693)/(t_(1/2))`

`k=\frac{0.693}{5730\text{ years}}`

`k=1.21* 10^(-4)\text{ years}^(-1)`

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`1.21* 10^(-4)\text{ years}^(-1)`

t = time passed by the sample = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = 100 - 25 = 75 g

Now put all the given values in above equation, we get

`t=(2.303)/(1.21* 10^(-4))\log(100)/(75)`

`t=2377.9\text{ years}=2.4* 10^3\text{ years}`

Therefore, the time passed by the sample is
`2.4* 10^3\text{ years}`

Part 2 :

Now we have to calculate the amount left.

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`1.21* 10^(-4)\text{ years}^(-1)`

t = time passed by the sample = 2255 years

a = let initial amount of the reactant = 1.5 mmol

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

`2255=(2.303)/(1.21* 10^(-4))\log(1.5)/(a-x)`

`a-x=1.14mmol`

Therefore, the amount left after decay process is 1.14 mmol.