How many atoms of `0.1 g`-atom of a radioacitve isotope `._(Z)X^(A)` (half = 5 days) will decay during the 11th day?

Explanation

Amount of radioactive substance = 0.1 g-atom
So `N_(0) = 0.1 xx` Avogadro's number
`= 0.1 xx 6.02 xx 10^(23)`
`= 6.02 xx 10^(22)` atoms
Number of atoms after 5 days `= (6.02 xx 10^(22))/(2) = 3.01 xx 10^(22)`
Number of atomsd after 10 days `= (3.01 xx 10^(22))/(2)`
`= 1.505 xx 10^(22)`
Let the number of atoms left after 11 days be `N`.
We konw that
`t = (2.303)/(K) log (N_(0))/(N)`
Given `t = 11, K = (0.693)/(5) , N_(0) = 6.02 xx 10^(22)`
So, `11 = (2.303 xx 5)/(0.693) log (6.02 xx 10^(22))/(N)`
or `log (6.02 xx 10^(22))/(N) = (11 xx 0.693)/(2.303 xx 5) = 0.6620`
`(6.02 xx 10^(22))/(N)` = Atomic 0.6620 = 4.592
So, `N = (6.02)/(4.592) xx 10^(22) = 1.3109 xx 10^(22)` atoms decayed during 11th days.