$(a)\;0.54 Ci\qquad(b)\;3.2 Ci \qquad(c)\;1.2 Ci\qquad(d)\;4.8 Ci$

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Answer : (a) $\;0.54 Ci$

Explanation :

Let $\;A_{0}=$ initial activity of sample

$A_{10}=$initial activity of isotope 1 &

$A_{20}=$ initial activity of isotope 2

$A_{0}=A_{10}+A_{20}$

Similarly for final activity (Activity after time t)

$A_{t}=A_{2t}+A_{1t}$

$A_{1t}=A_{10}\;e^{-\lambda_{1}t}+A_{20}e^{-\lambda_{2}t}$

Now in given equation

$A_{0}=2 \mu Ci \to \;A_{0}=A_{10}+A_{20}---(1)$

Initial ratio of atoms of isotopes = $\;2:1$

We know from definition of activity ,

$A=\lambda N$

$\large\frac{A_{10}}{A_{20}}=\large\frac{\lambda_{1} N_{10}}{\lambda_{2} N_{20}}=\large\frac{N_{10}}{N_{20}}\times\large\frac{T_{2}}{T_{1}}$

Where T represents half life

$\large\frac{A_{10}}{A_{20}}=\large\frac{2}{1}\times\large\frac{25}{14}=\large\frac{50}{14}=\large\frac{25}{7}$

On solving (1) & (2) we get $\;A_{20}=\large\frac{7}{16}\;$& $\;A_{10}=\large\frac{25}{16}$

$A_{t}=A_{10}e^{-\lambda_{1} t}+A_{20} e^{-\lambda_{2}t}$

$A_{1}=\large\frac{25}{16}\;e^{-\large\frac{0.693}{14}\times30}\times\large\frac{7}{10}\;e^{-\large\frac{0.693}{25}\times30}$

Consider $\;1^{st}\;$ exponential term $\;\large\frac{0.693\times3}{14}=e^{-1.483}$

Let $\;y=e^{-1.483}$

$ln y=-1.483\;$

$log\;y=-\large\frac{1.483}{2.303}\;\to y=antilog(-\large\frac{1.483}{2.303})$

So $\;e^{-x} =antilog(-\large\frac{-x}{2.303})$

At $=\large\frac{25}{16}\times0.2265+\large\frac{7}{16}\times0.4353$

$=0.54 Ci$

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