Ameteorite 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?

Answer : The number of half-lives passed since the meteorite formed is, 5

Explanation : Given,

Mass of nickel-59 = 0.17 g

Mass of cobalt-59 = 5.27 g

Equation for the radioactive decay of nickel-59 is :

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 :

Moles of

Moles of

By stoichiometry of the reaction,

1 mole of is produced by 1 mole

So, 0.089 moles of will be produced by =

Amount of decomposed will be = 0.089 moles

Initial amount of 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:

where,

a = amount of reactant left after n-half lives = 0.00288 moles

= Initial amount of the reactant = 0.09188 moles

n = number of half lives

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

Taking log on both sides, we get

Therefore, the number of half-lives passed since the meteorite formed is, 5

where are the vectors ?

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