Question

1010 years. What is the age of a rock sample that contains 1042 ug Rubidum 87 decays by β production to strontium-87 with a half life of 4.7 87Rb and 4.3 μg 87Sr ? years old Assume that no 87Sr was present when the rock was fomed The atomic masses are given in the table below: Nuclide Atomic Mass 87Rb 86.90919 87Sr 86.90888

Answer 1

Apply half life equation

A = A0*(1/2)^(t/hl)

A = A0 *(1/2)^(t/(4.7*10^10))

the decay:

87Rb = 87Sr + beta

mol of Sr = 4.3/69.90888 = 0.06150863

mol of Rb = 104.2/86.90919 = 1.19895

total mol = 1.19895+0.06150863 = 1.260459

now..

substitute

1.19895 = 1.260459*(1/2)^(t/(4.7*10^10))

solve for t

ln(1.19895 /1.260459) / ln(0.5) = t / (4.7*10^10)

t =  (4.7)*ln(1.19895 /1.260459) / ln(0.5) * 10^10

t = 0.33923*10^10

then

t = 3.39*10^9 years old