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 \(_{82}^{212}\textrm{Pb}\) radioisotope follows:
\(_{82}^{212}\textrm{Pb}\rightarrow _{83}^{212}\textrm{Bi}+_{-1}^0\beta\)
To calculate the initial amount of \(_{82}^{212}\textrm{Pb}\), we will require the stoichiometry of the reaction and the moles of the reactant and product.
Expression for calculating the moles is given by:
\(\text{no of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\)
Moles of \(_{82}^{212}\textrm{Pb}\) left = \(\frac{10.5g}{212g/mol}=0.0495moles\)
Moles of \(_{83}^{212}\textrm{Bi}=\frac{157.5g}{212g/mol}=0.7429moles\)
By the stoichiometry of above reaction,
1 mole of \(_{83}^{212}\textrm{Bi}\) is produced by 1 mole \(_{82}^{212}\textrm{Pb}\)
So, 0.7429 moles of \(_{83}^{212}\textrm{Bi}\) will be produced by = \(\frac{1}{1}\times 0.7429=0.7429\text{ moles of }_{82}^{212}\textrm{Pb}\)
Amount of \(_{82}^{212}\textrm{Pb}\) decomposed will be = 0.7429 moles
Initial amount of \(_{82}^{212}\textrm{Pb}\) 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:
\(a=\frac{a_o}{2^n}\)
where,
a = amount of reactant left after n-half lives = 0.0495 moles
\(a_o\) = Initial amount of the reactant = 0.7924 moles
n = number of half lives
Putting values in above equation, we get:
\(0.0495=\frac{0.7924}{2^n}\)
\(2^n=16.0080\)
Taking log on both sides, we get
\(n\log2=\log(16.0080)\
=4\)
