Half - life for radioactive `.^(14)C` is 5760 years. In how many years 200 mg of `.^(14)C` will be reduced to 25 mg ?

Options

A. 11520 years
B. 23040 years
C. 5760 years
D. 17280 years

Answer

D

Explanation

`25 = [(1)/(2)]^(n) xx 200, [(1)/(2)]^(n) = (25)/(200) = (1)/(8) = [(1)/(2)]^(3)`
n = 3, Number of half lives = 3
so time required `= 3xx 5760 = 17280 yr`