Answer: The sample must have passed 4 half-lives after the sample was originally formed.
Explanation: This is a type of radioactive decay and all the radioactive process follow first order kinetics.
Equation for the reaction of decay of [tex]_{82}^{212}\textrm{Pb}[/tex] radioisotope follows:
[tex]_{82}^{212}\textrm{Pb}\rightarrow _{83}^{212}\textrm{Bi}+_{-1}^0\beta[/tex]
To calculate the initial amount of [tex]_{82}^{212}\textrm{Pb}[/tex], we will require the stoichiometry of the reaction and the moles of the reactant and product.
Expression for calculating the moles is given by:
[tex]\text{no of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of [tex]_{82}^{212}\textrm{Pb}[/tex] left = [tex]\frac{10.5g}{212g/mol}=0.0495moles[/tex]
Moles of [tex]_{83}^{212}\textrm{Bi}=\frac{157.5g}{212g/mol}=0.7429moles[/tex]
By the stoichiometry of above reaction,
1 mole of [tex]_{83}^{212}\textrm{Bi}[/tex] is produced by 1 mole [tex]_{82}^{212}\textrm{Pb}[/tex]
So, 0.7429 moles of [tex]_{83}^{212}\textrm{Bi}[/tex] will be produced by = [tex]\frac{1}{1}\times 0.7429=0.7429\text{ moles of }_{82}^{212}\textrm{Pb}[/tex]
Amount of [tex]_{82}^{212}\textrm{Pb}[/tex] decomposed will be = 0.7429 moles
Initial amount of [tex]_{82}^{212}\textrm{Pb}[/tex] will be = Amount decomposed + Amount left = (0.0495 + 0.7429)moles = 0.7924 moles
Now, to calculate the number of half lives, we use the formula:
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives = 0.0495 moles
[tex]a_o[/tex] = Initial amount of the reactant = 0.7924 moles
n = number of half lives
Putting values in above equation, we get:
[tex]0.0495=\frac{0.7924}{2^n}[/tex]
[tex]2^n=16.0080[/tex]
Taking log on both sides, we get
[tex]n\log2=\log(16.0080)\\n=4[/tex]
