Correct Answer - d
`k = (0.693)/(t_(1//2)) = (0.693)/(10 "years")`
If initial concentration a = 10 gm and final concentration
`x = (a)/(2) = 5` gm
then , t = `(2.303)/(k)` log `(a)/(a-x) = (2.303)/(0.693) xx 10 xx "log" (10)/(5)`
`= (2.303 xx 10 xx "log" 2)/(0.693) = (2.303 xx 10 xx 0.301)/(0.693) = 10 ` years .