Answer:
option A
Step-by-step explanation:
given,
sample of tritium, ³H
[tex]t_{1/2} = 12.32 years[/tex]
contents present of ³H = 5.25 mol
content of ³He = 6.35 mol
reaction
[tex]^3H\rightarrow \ ^3He[/tex]
A₀ is the initial concentration
At is the concentration after time t
A₀ = 5.25 + 6.35 = 11.6 mol
At = 5.25
now,
[tex]k = \dfrac{0.693}{t_{1/2}}[/tex]
[tex]k = \dfrac{0.693}{12.32}[/tex]
k = 0.0563 /year
[tex]ln(\dfrac{A_o}{A_t}) = k t[/tex]
[tex]t = \dfrac{1}{k} ln(\dfrac{A_o}{A_t})[/tex]
[tex]t = \dfrac{1}{0.0563} ln(\dfrac{11.6}{5.25})[/tex]
t = 14.1 yr
hence, the correct answer is option A