Answer:
4 half-lives will occur during this period of time.
Explanation:
Formula used :
[tex]a_o=a\times e^{-\lambda t}[/tex]
[tex]\lambda =\frac{0.693}{t_{1/2}}[/tex]
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives and time t
[tex]a_o[/tex] = Initial amount of the reactant.
[tex]\lambda = [/tex] decay constant
[tex]t_{1\2}[/tex] = half life of an isotope
n = number of half lives
We have :
[tex]a_o=100.0 g[/tex]
a = ?
t = 552 days
[tex]t_{1/2}=138 days[/tex]
[tex]a=100.0 g\times e^{-\frac{0.693}{138}\times 552}[/tex]
[tex]a=6.254 g[/tex]
[tex]6.254 g=\frac{100.0 g}{2^n}[/tex]
[tex]2^n=\frac{100.0 g}{6.254 g}[/tex]
n = 4
4 half-lives will occur during this period of time.
