# Problem: Rubidium-87 decays by β-particle production to strontium-87 with a half-life of 4.7 x 1010 years. What is the age of a rock sample that contains 109.7 μg of  87Rb and 3.1 μg of  87Sr? Assume that no 87Sr was present when the rock was formed. The atomic masses for   87Rb and 87Sr are 86.90919 u and 86.90888 u, respectively.

###### FREE Expert Solution

Recall that radioactive/nuclear decay of isotopes follows first-order kinetics, and the integrated rate law for first-order reactions is:

where:

[N]t = concentration at time t

k = decay constant

t = time

[N]0 = initial concentration

Also, recall that half-life is the time needed for the amount of a reactant to decrease by 50% or one-half

The half-life of a first-order reaction is given by:

We first need to calculate for the decay constant using the given half-life of 87Rb, 4.7 x 1010 years:

k = 1.4748x10-11 yr-1

Calculate the initial amount of 87Rb atoms present:

initial amount of 87Rb = final amount of 87Rb + amount of 87Sr

amu is approximately equal to molar mass (g/mol)

87Rb = 7.6012x1017 atoms

87Sr = 2.148x1016 atoms

initial amount of 87Rb = 7.6012x1017 atoms + 7.6012x1017 atoms

initial amount of 87Rb = 7.816x1017 atoms

Calculate the age of a rock sample:

[N]0 =  87Rb initial = 7.816x1017 atoms                        k = 1.4748x10-11 yr–1

[N]t = 87Rb final = 7.6012x1017 atoms                         t = ???

Solving for t:

97% (465 ratings) ###### Problem Details

Rubidium-87 decays by β-particle production to strontium-87 with a half-life of 4.7 x 1010 years. What is the age of a rock sample that contains 109.7 μg of  87Rb and 3.1 μg of  87Sr? Assume that no 87Sr was present when the rock was formed. The atomic masses for   87Rb and 87Sr are 86.90919 u and 86.90888 u, respectively.