Answer:
The age of the alloy is 73.94 years.
Explanation:
For metal A:
Initial amount of metal A = x
Final amount after time t = 0.53 kg
Half life of metal A =[tex]t_{1/2}=12 years[/tex]
[tex]\lambda =\frac{0.693}{12 years}=0.05775 year^{-1}[/tex]
[tex]0.53=x\times e^{-0.05775 year^{-1} t}[/tex]..(1)
For metal B:
Initial amount of metal B = x
Final amount after time t = 2.20 kg
Half life of metal B =[tex]t_{1/2}=18 years[/tex]
[tex]\lambda '=\frac{0.693}{18 years}=0.0385 year^{-1}[/tex]
[tex]2.20=x\times e^{-0.0385 year^{-1} t}[/tex]..(2)
Dividing (1) by (2)
[tex]\frac{0.53}{2.20}=\frac{x\times e^{-0.05775 year^{-1} t}}{x\times e^{-0.0385 year^{-1} t}}[/tex]
[tex]0.2409=e^{t(0.0385-0.05775)}[/tex]
[tex]\ln 0.2409=t(-0.01925)[/tex]
t = 73.94 years