#### Question

A sample contains a mixture of ^{108}Ag and ^{110}Ag isotopes each having an activity of 8.0 × 10^{8} disintegration per second. ^{110}Ag is known to have larger half-life than ^{108}Ag. The activity A is measured as a function of time and the following data are obtained.

Time (s) |
Activity (A) (10 ^{8} disinte-grations s ^{−1}) |
Time (s) |
Activity (A 10 ^{8} 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/A_{0}) versus time. (b) See that for large values of time, the plot is nearly linear. Deduce the half-life of ^{110}Ag from this portion of the plot. (c) Use the half-life of ^{110}Ag to calculate the activity corresponding to ^{108}Ag in the first 50 s. (d) Plot In (A/A_{0}) versus time for ^{108}Ag for the first 50 s. (e) Find the half-life of ^{108}Ag.

#### Solution

(a) Activity, A_{0} = 8 × 10^{8} dis/sec

(i) `"In" (A_1/A_0) = "In" (11.794/8) = 0.389`

(ii) `"In" (A_2/A_0) = "In" (9.1680/8) = 0.12362`

(iii) `"In" (A_3/A_0) = "In" (7.4492/8) = -0.072`

(iv) `"In" (A_4/A_0) = "In" (6.2684/6) = -0.244`

(v) `"In" (A_5/A) = "In" (5.4115/8) = -0.391`

(vi) `"In" (A_6/A_0) = "In" (3.0828/8) = -0.954`

(vii) `"In" (A_7/A_0) = "In" (91.8899/8) = -1.443`

(viii) `"In" (1.1671/8) = "In" (90.7212/8) = -1.93`

(ix) `"In" (0.7212/8) = "In" (90.7212/8) = -2.406`

The required graph is given below.

(b) Half-life of ^{110}Ag = 24.4 s

(c) Half-life of ^{110}Ag, `T_"1/2"`= 24.4 s

Decay constant, `lambda = 0.693/T_"1/2"`

⇒ `lambda = 0.63/24.4 = 0.0284`

`therefore t = 50 sec`

Activity , `"A" = "A"_0e^(-lambdat)`

= `8 xx 10^8 xx e^(-0.0284 xx 50)`

= `1.93 xx 10^8`

(d)

(e) The half-life period of `""^108"Ag"` that you can easily watch in your graph is 144 s .