Respuesta :
Answer: 3
Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.
Equation for the reaction of decay of [tex]_{19}^{40}\textrm{K}[/tex] radioisotope follows:
[tex]Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles[/tex]
[tex]Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles[/tex]
[tex]_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e[/tex]
By the stoichiometry of above reaction,
1 mole of [tex]_{40}^{91}\textrm{Zr}[/tex] is produced by 1 mole [tex]_{41}^{91}\textrm{Nb}[/tex]
So, 0.17 moles of [tex]_{40}^{91}\textrm{Zr}[/tex] will be produced by = [tex]\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}[/tex]
Amount of [tex]_{82}^{212}\textrm{K}[/tex]
decomposed will be = 0.17 moles
Initial amount of [tex]_{40}^{91}\textrm{Nb}[/tex] will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives = 0.024
[tex]a_o[/tex] = Initial amount of the reactant = 0.194
n = number of half lives= ?
Putting values in above equation, we get:
[tex]0.024=\frac{0.194}{2^n}[/tex]
[tex]n=3[/tex]
Therefore, 3 half lives have passed.
