I took the liberty of finding for the complete question.
And here I believe that the problem asks for the half life of Curium. Assuming
that the radioactive decay of Curium is of 1st order, therefore the
rate equation is in the form of:
A = Ao e^(-kt)
where,
A = amount after t years = 2755
Ao = initial amount = 3312
k = rate constant
t = number of years passed = 6
Therefore the rate constant is:
2755/3312 = e^(-6k)
-6k = ln (2755/3312)
k = 0.0307/yr
The half life, t’, can be calculated using the formula:
t’ = ln 2 / k
Substituting the value of k:
t’ = ln 2 / 0.0307
t’ = 22.586 years
or
t’ = 22.6 years
