Half life of `.^(24)Na` is 14.8 hours. In what period of time will a sample of this element lose 90% of its activity?

Explanation

Let the initial radioactive intensity be 100 which corresponds to `N^(0)`. The radioactive intensity after a time period, say t hours, will be 10 (corresponding to N) as the substance has lost 90% of its radioactive intensity.
`lamda= (0.6932)/(t_((1)/(2))) = (0.6932)/(14.8)`
We have, `lamda= (2.303)/(t) "log" (N^(0))/(N)`
`(0.6932)/(14.8) =(2.303)/(t) "log" (100)/(10)`
t= 49.17 hours.