Question

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives have passed since the sample originally formed?

Answer 1

3 half lives

Hope I helped :>

Answer 2

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of
`_(19)^(40)\textrm{K}` radioisotope follows:

`Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=(15.4)/(91)=0.17moles`

`Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=(2.2)/(91)=0.024moles`

`_(41)^(91)\textrm{Nb}\rightarrow _(40)^(91)\textrm{Zr}+_(+1)^0e`

By the stoichiometry of above reaction,

1 mole of
`_(40)^(91)\textrm{Zr}` is produced by 1 mole
`_(41)^(91)\textrm{Nb}`

So, 0.17 moles of
`_(40)^(91)\textrm{Zr}` will be produced by =
`(1)/(1)* 0.17=0.17\text{ moles of }_(40)^(91)\textrm{Nb}`

Amount of
`_(82)^(212)\textrm{K}`

decomposed will be = 0.17 moles

Initial amount of
`_(40)^(91)\textrm{Nb}` will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

`a=(a_o)/(2^n)`

where,

a = amount of reactant left after n-half lives = 0.024

`a_o` = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

`0.024=(0.194)/(2^n)`

`n=3`

Therefore, 3 half lives have passed.