Answer: The sample must have passed 4 half-lives after the sample was originally formed.
Explanation: This is a type of radioactive decay and all the radioactive process follow first order kinetics.
Equation for the reaction of decay of
radioisotope follows:
![_{82}^{212}\textrm{Pb}\rightarrow _{83}^{212}\textrm{Bi}+_{-1}^0\beta](https://tex.z-dn.net/?f=_%7B82%7D%5E%7B212%7D%5Ctextrm%7BPb%7D%5Crightarrow%20_%7B83%7D%5E%7B212%7D%5Ctextrm%7BBi%7D%2B_%7B-1%7D%5E0%5Cbeta)
To calculate the initial amount of
, we will require the stoichiometry of the reaction and the moles of the reactant and product.
Expression for calculating the moles is given by:
![\text{no of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7Bno%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Moles of
left =
Moles of ![_{83}^{212}\textrm{Bi}=\frac{157.5g}{212g/mol}=0.7429moles](https://tex.z-dn.net/?f=_%7B83%7D%5E%7B212%7D%5Ctextrm%7BBi%7D%3D%5Cfrac%7B157.5g%7D%7B212g%2Fmol%7D%3D0.7429moles)
By the stoichiometry of above reaction,
1 mole of
is produced by 1 mole ![_{82}^{212}\textrm{Pb}](https://tex.z-dn.net/?f=_%7B82%7D%5E%7B212%7D%5Ctextrm%7BPb%7D)
So, 0.7429 moles of
will be produced by = ![\frac{1}{1}\times 0.7429=0.7429\text{ moles of }_{82}^{212}\textrm{Pb}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%5Ctimes%200.7429%3D0.7429%5Ctext%7B%20moles%20of%20%7D_%7B82%7D%5E%7B212%7D%5Ctextrm%7BPb%7D)
Amount of
decomposed will be = 0.7429 moles
Initial amount of
will be = Amount decomposed + Amount left = (0.0495 + 0.7429)moles = 0.7924 moles
Now, to calculate the number of half lives, we use the formula:
![a=\frac{a_o}{2^n}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Ba_o%7D%7B2%5En%7D)
where,
a = amount of reactant left after n-half lives = 0.0495 moles
= Initial amount of the reactant = 0.7924 moles
n = number of half lives
Putting values in above equation, we get:
![0.0495=\frac{0.7924}{2^n}](https://tex.z-dn.net/?f=0.0495%3D%5Cfrac%7B0.7924%7D%7B2%5En%7D)
![2^n=16.0080](https://tex.z-dn.net/?f=2%5En%3D16.0080)
Taking log on both sides, we get
![n\log2=\log(16.0080)\\n=4](https://tex.z-dn.net/?f=n%5Clog2%3D%5Clog%2816.0080%29%5C%5Cn%3D4)