The half-life of stronitum-90 is 28 years. How many grams of this nuclide in a 4.00g sample will remain after 112 years?
k = 0.693/t1/2 = 0.693/28 = ??
Substitute for k in the equation below.
ln(No/N) = kt
ln(4.00/N) = k(112)
Solve for N.
Substitute for k in the equation below.
ln(No/N) = kt
ln(4.00/N) = k(112)
Solve for N.