A radioactive sample has `2.6 mu g` of pure `""_(7)^(13)"N"` which has a half - life of 10 minutes.
(a) How many nuclei are present initially?
(b) What is the activity initially?
(c) What is the activity after 2 hours?
(d) Calculate mean life of this sample.

Explanation

(a) To find `N_(0)` we have to find the number of `""_(7)^(13)"N"` atoms is `2.6 mu g`. The atomic mass of nitrogen is 13. Therefore, 13 g of `""_(7)^(13)"N"` contains Avogadro number `(6.02 xx 10^(23))` of atoms.
In 1 g, the number of `""_(7)^(13)"N"` is equal to be `(6.02 xx 10^(23))/(13)` atoms. So the numberof `""_(7)^(13)"N"` atoms in `2.6 mug` is
`N_(0) = (6.02 xx 10^(23))/(13) xx 2.6 xx 10^(-6) = 12.04 xx 10^(6)` atoms.
(b) To find the initial activity `R_(0)`, we have to evaluate decay constant `lambda`
`lambda = (0.6931)/(T_(1/2)) = (0.6931)/(10 xx 60) = 1.155 xx 10^(-3) s^(-1)`
Therefore
`R_(0) = lambda N_(0) = 1.155 xx 10^(-3) xx 12.04 xx 10^(16)`
`= 13.90 xx 10^(13) (decays)/(s)`
`= 13.90 xx 10^(13)Bq`
In terms or a curie, `R_(0) = (13.90 xx 10^(13))/(3.7 xx 10^(10)) = 3.75 xx 10^(3)Ci`
since 1Ci = `3.7 xx 10^(10)Bq`
(c) Activity after 2 hours can be calculated in two different ways:
Methid 1 : `R = R_(0)e^(-lambda t)`
At t = 2 hr = 7200 s
`R = 3.75 xx 10^(3) xx e^(-7200) xx 1.155 xx 10^(-3)`
`R = 3.75 xx 10^(3) xx 2.4 xx 10^(-4) = 0.9 Ci`
Method 2 : `R = ((1)/(2))^(n) R_(0)`
Here `n = (120 min)/(10 min) = 12`
` R = ((1)/(2))^(12) xx 3.75 xx 10^(3) approx 0.9 Ci`
(d) mean life `tau = (T_(1/2))/(0.6931) = (10 xx60)/(0.6931) = 865.67 s`