Question

A sample contains 10.5 g of the radioisotope Pb-212 and 157.5 g of its daughter isotope, Bi-212. How many half-lives have passed since the sample originally formed?
4
14
15
147.5

Answer 1

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

$n\log2=\log(16.0080)\\n=4$