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

(a) Activity, A₀ = 8 × 10⁸ dis/sec

(i) ${\displaystyle \text{In}\left(\frac{A_1}{A_0}\right)=\text{In}\left(\frac{11.794}{8}\right)=0.389}$

(ii) ${\displaystyle \text{In}\left(\frac{A_2}{A_0}\right)=\text{In}\left(\frac{9.1680}{8}\right)=0.12362}$

(iii) ${\displaystyle \text{In}\left(\frac{A_3}{A_0}\right)=\text{In}\left(\frac{7.4492}{8}\right)=-0.072}$

(iv) ${\displaystyle \text{In}\left(\frac{A_4}{A_0}\right)=\text{In}\left(\frac{6.2684}{6}\right)=-0.244}$

(v) ${\displaystyle \text{In}\left(\frac{A_5}{A}\right)=\text{In}\left(\frac{5.4115}{8}\right)=-0.391}$

(vi) ${\displaystyle \text{In}\left(\frac{A_6}{A_0}\right)=\text{In}\left(\frac{3.0828}{8}\right)=-0.954}$

(vii) ${\displaystyle \text{In}\left(\frac{A_7}{A_0}\right)=\text{In}\left(\frac{91.8899}{8}\right)=-1.443}$

(viii) ${\displaystyle \text{In}\left(\frac{1.1671}{8}\right)=\text{In}\left(\frac{90.7212}{8}\right)=-1.93}$

(ix) ${\displaystyle \text{In}\left(\frac{0.7212}{8}\right)=\text{In}\left(\frac{90.7212}{8}\right)=-2.406}$

The required graph is given below.

(b) Half-life of ¹¹⁰Ag = 24.4 s
(c) Half-life of ¹¹⁰Ag,  ${\displaystyle T_{\text{1/2}}}$= 24.4 s

Decay constant, ${\displaystyle \lambda =\frac{0.693}{T_{\text{1/2}}}}$

⇒ ${\displaystyle \lambda =\frac{0.63}{24.4}=0.0284}$

${\displaystyle \therefore t=50\sec }$

Activity , ${\displaystyle \text{A}={\text{A}}_0{e}^{-\lambda t}}$

= ${\displaystyle 8\times {10}^8\times e^{-0.0284\times 50}}$

= ${\displaystyle 1.93\times {10}^8}$

(d)

(e) The half-life period of ${\displaystyle {}^{108}\text{Ag}}$ that you can easily watch in your graph is 144 s .