How long will it take for 30% of the c-14 atoms in a sample of c-14 to decay?

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

Leave a Comment