Question

A sample contains a mixture of 108Ag and 110Ag isotopes each having an activity of 8.0 × 108 disintegration per second. 110Ag is known to have larger half-life than 108Ag. The activity A is measured as a function of time and the following data are obtained.
Time (s)
Activity (A)
(108 disinte-
grations s−1)
Time (s)
Activity (A
108 disinte-grations s−1)
20
40
60
80
100
11.799
9.1680
7.4492
6.2684
5.4115
200
300
400
500
3.0828
1.8899
1.1671
0.7212
(a) Plot ln (A/A0) versus time. (b) See that for large values of time, the plot is nearly linear. Deduce the half-life of 110Ag from this portion of the plot. (c) Use the half-life of 110Ag to calculate the activity corresponding to 108Ag in the first 50 s. (d) Plot In (A/A0) versus time for 108Ag for the first 50 s. (e) Find the half-life of 108Ag.

Answer 1

Answer:

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Answer 2

Explanation:

(a) The activity, $A_{0} \text { is } 8 \times 10^{8} \text { dis/sec. }$

(i) InA1A0 = 0.389

InA2A0 = 0.12362

InA3A0 = -0.072

InA4A0 = -0.244

InA5A = -0.391

InA6A0 = -0.954

InA7A0 = In91.88998 = -1.443

In1.16718 = -1.93

In0.72128 = -2.406

The essential graph is shown below.  

(b) 110Ag has the half-life = 24.4 s

(c) 110Ag has the half-life, T12 = 24.4 s

Constant of decay, λ = 0.693T12

⇒ λ = 0.0284

Therefore, t = 50 sec  

Activity, A = A0e-λt                  

$=1.93 \times 108(\mathrm{d})$

(e) 108Ag has the half-life period, which can be easily seen in the graph. It is 144 s.