Question

A sealed box was found which stated to have contained an alloy composed of equal parts by weight of two metals A and B. These met als are radioactive, with half lives of 12 years and 18 years, respectively and when the container was opened it was found to contain 0.53 kg of A and 2.20 kg of B. Deduce the age of the alloy.

Answer 1

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 =$t_{1/2}=12 years$

$\lambda =\frac{0.693}{12 years}=0.05775 year^{-1}$

$0.53=x\times e^{-0.05775 year^{-1} t}$..(1)

For metal B:

Initial amount of metal B = x

Final amount after time t = 2.20 kg

Half life of metal B =$t_{1/2}=18 years$

$\lambda '=\frac{0.693}{18 years}=0.0385 year^{-1}$

$2.20=x\times e^{-0.0385 year^{-1} t}$..(2)

Dividing (1) by (2)

$\frac{0.53}{2.20}=\frac{x\times e^{-0.05775 year^{-1} t}}{x\times e^{-0.0385 year^{-1} t}}$

$0.2409=e^{t(0.0385-0.05775)}$

$\ln 0.2409=t(-0.01925)$

t = 73.94 years