Respuesta :
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 :
[tex]_{28}^{59}\textrm{Ni}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^{0}\beta[/tex]
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 : [tex]\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of [tex]_{28}^{59}\textrm{Ni}=\frac{0.17g}{59g/mol}=0.00288moles[/tex]
Moles of [tex]_{27}^{59}\textrm{Co}=\frac{5.27g}{59g/mol}=0.089moles[/tex]
By stoichiometry of the reaction,
1 mole of [tex]_{27}^{59}\textrm{Co}[/tex] is produced by 1 mole [tex]_{28}^{59}\textrm{Ni}[/tex]
So, 0.089 moles of [tex]_{27}^{59}\textrm{Co}[/tex] will be produced by = [tex]\frac{1}{1}\times 0.089=0.089\text{ moles of }_{28}^{59}\textrm{Ni}[/tex]
Amount of [tex]_{28}^{59}\textrm{Ni}[/tex] decomposed will be = 0.089 moles
Initial amount of [tex]_{28}^{59}\textrm{Ni}[/tex] 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 :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives = 0.00288 moles
[tex]a_o[/tex] = Initial amount of the reactant = 0.09188 moles
n = number of half lives
Now put all the given values in above equation, we get
[tex]0.00288=\frac{0.09188}{2^n}[/tex]
[tex]2^n=32.81[/tex]
Taking log on both sides, we get
[tex]n\log2=\log(32.81)\\n=5.03=5[/tex]
Therefore, '5' number of half-lives have passed since the meteorite formed .
