Problema Solution
If 12g of radioactive substance is present initially, and 5 years later only 6g remain, how much of the substance will be present after 7 years?
Answer provided by our tutors
we will use the formula
A=Pe^(rt)
P = 12 g
t = 5 tears
A = 6g
r = ?
6= 12 e^(4r)
e^4r= 6/12
ln (e^4r) = ln 0.5
r= (1/4)ln 0.5
r=-0.1733
the formula is
A=Pe^(-0.1733t)
for t = 7 years
A= 12 e^(7(-0.1733))
A = 3.5676 gr
after 7 years there will be 3.5676 grams present.