Question

The half-life for the radioactive decay of C−14 is 5730 years.

Part A: How long will it take for 25% of the C−14 atoms in a sample of C−14 to decay?

Part B: If a sample of C−14 initially contains 1.9 mmol of C−14, how many millimoles will be left after 2255 years?

Answer 1

Part orA: For a first order reaction rate constant ,

k = ( 2.303 /t )x log ( Mo / M)
Where
Mo = initial mass/atoms
M= Mass/atoms left after time t = Mo-(25%Mo)= 75%Mo
t = time = ?

k= rate constant= 0.693/half-life= 0.693/5730 years

k= 1.21*10^-4 yr^-1

Plug the values we get t= (2.303/k)*log(Mo/M)

t= 2379 yr

Part B:

For this Mo= 1.9 mmol

M= ?

t= 2255 years

k= 1.21*10^-4 yr^-1

So log(Mo/M)= (kt)/2.303

= (1.21*10^-4 yr^-1 *2255 yr)/2.303= 0.1184

Mo/M= 10^0.1184= 1.314

M= Mo/1.314

= 1.9mmol/1.314

= 1.4 mmol