The half-life for the radioactive decay of C?14 is 5730

years. How long will it take for 30% of the C?14 atoms in a sample of

C?14 to decay? Express your answer using two significant figures. If a sample of C?14 initially contains 1.9mmol ofC?14, how many

millimoles will be left after 2255 years? Express your answer using two significant figures.

Hi,

The amount A remaining from an original sample Ao after time t

has passed is given by A = Aoe^(-kt)

t1/2 = 5730 years

decay constant k = ln 2/5730

if 30% decays then 70% remains

.7 = e^(-kt)

ln(0.7) = -kt

t = -ln(0.7)*5730/ln2

t = 2948 years for 30% to decay

A = 1.9 *e^(-k*2255)

A = 1.9*0.7613

A = 1.45 mmol will remain after 2255

years

One form of the first order rate equation is-

ln(Ao/A) = kt

where Ao = the initial concentration and A = the concentration

after time t. k = the rate constant.

If t = the half life, then Ao/A = 2 and ln(2) = 0.693

0.693 = k * 5730 yr

k = 1.210×10^-4 yr^-1

If 30% decays, 70% is left

ln(100/70) = 1.210×10^-4 * t

t = 2949 year

ln(1.9 mmol/X mmol) = 1.210×10^-4 x 2285

ln(1.9/X) = 0.2764

1.9/X = 1.318

X = 1.9/1.318 = 1.44 mmol1.44 mmol left after 2285 years