Question

In: Chemistry

Rubidium-87 decays by β production to strontium-87 with a half life of 4.7 × 1010 years....

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

Solutions

Expert Solution

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


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