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A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answer :

Mickymike92go

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

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