A sample contains a mixture of `Ag^110` and `Ag^108` isotopes each having an activity of `8.0xx10^8 `disintergrations per second. `Ag^108` is known to have larger half-life than `Ag^110`.The activity A is measured as a fuction of time and the following data are obtained .

(a) plot in `(A//A_0)` versus time. (b) See that for large values of time, the plot is nearly linear .Deduce the half-life of `Ag^108` from this portion of the plot .(c ) use the half-life of `Ag^108` to calculate the activity corresponding to `Ag^110` in the first 50 s . (d) Plot In `(A//A_0)`versus time for `Ag^110` for the first 50 s. (e) find the half-life of `Ag^110`.

Answer

A, B, D

Explanation

Activities of sample containing `_(108)Ag` and
`_(110)Ag` isotopes`=8.0 xx 10^8 dis//sec.`
(a) Here we take A `= 8 xx 10^8 dis//sec`
`:. (i) In (A_1/A_(01) = In (11.794/8) = 0.389`
`(ii) In (A_2/A_(02) = In (9.1680/8) = 0.12362`
`(iii) In (A_3/A_(03) = In (7.4492/8) = -0.072`
(iv) `In (A_2/A_(04) = In (6.2684/6) = -0.244`
(v) `In (5.4115/8) = -0.391`
(vi) `In (3.0828/8) = -0.954`
(vii) `In (91.8899/8) = -1.443`
(viii) `In (1.167/8) = -1.93`
(ix) `In (90.7212/8) = -2.406`
(b) The half life of `_(110)Ag` from his part of
the plot is `24.4 s`.
(c) Half life of `_(110)Ag = 24.4s`
`:. Decay constant, lambda = 0.693/24.4 = 0.0284`
`:. t = 50 sec`
The activity, `A = A_(0) e^(-lambdat)`
`= 8 xx 10^8 xx e^(-0.0284 xx 50)`
`= 1.93 xx 10^8`
(d) Plot yourself .
(e) The half life period of `_(108)Ag that you
can easilty watch in your graph is 144 s.