Question

A meteorite contains 0.17 g of nickel-59, a radioisotope that decays to form cobalt-59. The meteorite also contains 5.27 g of cobalt-59. How many nickel-59 half-lives have passed since the meteorite formed? 1 5 5.1 5.44

Answer 1

the answer is 5 to this question 

Answer 2

Answer : The correct option is, 5

Solution : Given,

Mass of nickel-59 = 0.17 g

Mass of cobalt-59 = 5.27 g

Equation for the radioactive decay of nickel-59 is :

`_(28)^(59)\textrm{Ni}\rightarrow _(27)^(59)\textrm{Co}+_(-1)^(0)\beta`

Now, we have to calculate the initial amount of nickel-59, we are using the stoichiometry of the reaction and moles of the reactant and product.

Formula used :
`\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of
`_(28)^(59)\textrm{Ni}=(0.17g)/(59g/mol)=0.00288moles`

Moles of
`_(27)^(59)\textrm{Co}=(5.27g)/(59g/mol)=0.089moles`

By stoichiometry of the reaction,

1 mole of
`_(27)^(59)\textrm{Co}` is produced by 1 mole
`_(28)^(59)\textrm{Ni}`

So, 0.089 moles of
`_(27)^(59)\textrm{Co}` will be produced by =
`(1)/(1)* 0.089=0.089\text{ moles of }_(28)^(59)\textrm{Ni}`

Amount of
`_(28)^(59)\textrm{Ni}` decomposed will be = 0.089 moles

Initial amount of
`_(28)^(59)\textrm{Ni}` will be = Amount decomposed + Amount left = (0.00288 + 0.089)moles = 0.09188 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 = 0.00288 moles

`a_o` = Initial amount of the reactant = 0.09188 moles

n = number of half lives

Now put all the given values in above equation, we get

`0.00288=(0.09188)/(2^n)`

`2^n=32.81`

Taking log on both sides, we get

`n\log2=\log(32.81)\\n=5.03=5`

Therefore, '5' number of half-lives have passed since the meteorite formed .