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

4

Step-by-step explanation:

Edg 2020

Answer 2

Answer: The sample must have passed 4 half-lives after the sample was originally formed.

Step-by-step 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 =
`(10.5g)/(212g/mol)=0.0495moles`

Moles of
`_(83)^(212)\textrm{Bi}=(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 =
`(1)/(1)* 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=(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=(0.7924)/(2^n)`

`2^n=16.0080`

Taking log on both sides, we get

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