Answer : The number of half-life periods will be, 3
Explanation : Given,
Initial amount of lead = 10 kg
Amount of lead after decay = 1.25 kg
Half-life = 22 years
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives
[tex]a_o[/tex] = Initial amount of the reactant
n = number of half lives
[tex]t_{1/2}[/tex] = half-life
Now put all the given values in the above formula, we get:
[tex]1.25=\frac{10}{2^n}[/tex]
[tex]2^n=8[/tex]
[tex]2^n=2^3[/tex]
[tex]n=3[/tex]
Thus, the number of half-life periods will be, 3
