The time required for the decompoistion of `99.9%` fraction of a first order reaction is………….to that of its half-life time.

Explanation

We know that, `t = (2.303)/(lambda) log, (N_(0))/(N)`
`N_(0) = 100, N = (100 - 99.9) = 0.1`
so, Time required for 99.9% decay, `t = (2.303)/(lambda) log. (100)/(0.1)`
`= (2.303)/(lambda) xx 3`
Half life period ` = (0.693)/(lambda)`
so, `("Time required for 99.9% decay")/("Half life period") = (2.303 xx 3)/(lambda) xx (lambda)/(0.693)`
= 10