In: Chemistry
Rubidium-87 decays by β production to strontium-87 with a half life of 4.7 × 1010 years. What is the age of a rock sample that contains 104.6 µg 87Rb and 7.1 µg 87Sr ?
Assume that no 87Sr was present when the rock was
formed.
The atomic masses are given below:
Nuclide | Atomic Mass |
87Rb | 86.90919 |
87Sr | 86.90888 |
Age of rock = 4.45 x 109 years
Explanation
mass 87Rb = 104.6 ug
moles 87Rb = (mass 87Rb) / (molar mass 87Rb)
moles 87Rb = (104.6 ug) / (86.90919 g/mol)
moles 87Rb = 1.204 umol
moles 87Sr = (mass 87Sr) / (molar mass 87Sr)
moles 87Sr = (7.1 ug) / (86.90888 g/mol)
moles 87Sr = 0.0817 umol
moles 87Rb decayed = moles 87Sr present
moles 87Rb decayed = 0.0817 umol
total moles 87Rb = moles 87Rb present + moles 87Sr decayed
total moles 87Rb = (1.204 umol) + (0.0817 umol)
total moles 87Rb = 1.285 umol
According to the radioactive decay law,
final amount = (initial amount) / (2a)
where a = number of half lives passed
final amount = 1.204 umol
initial amount = 1.285 umol
Substituting the values,
1.204 umol = (1.285 umol) / (2a)
2a = (1.285 umol) / (1.204 umol)
2a = 1.068
a = log(1.068) / log(2)
a = 0.0947
time passed = (half lives passed) * (half life)
time passed = (0.0947) * (4.7 x 1010 years)
time passed = 4.453 x 109 years