Question

A solution contains a mixture of isotopes of $X^{A_1}(t_{1/2}=14days)$ and $X^{A_2}(t_{1/2}=25days)$. Total activity is 1 curie at t = 0. The activity reduces by $50\%$ in 20 days. The initial activities of $X^{A_1}$ and $X^{A_2}$ are :

• A. Initial activity of $X^{A_1}=0.3669\times 3.7\times {10}^{10}dps$; initial activity of $X^{A_2}=0.6331\times 3.7\times {10}^{10}dps$
• B. Initial activity of $X^{A_1}=0.6331\times 3.7\times {10}^{10}dps$; initial activity of $X^{A_2}=0.6331\times 3.7\times {10}^{10}dps$
• C. Initial activity of $X^{A_1}=0.6331\times 3.7\times {10}^{10}dps$; initial activity of $X^{A_2}=0.3669\times 3.7\times {10}^{10}dps$
• D. none of these

Answer 1

The correct option is A Initial activity of $X^{A_1}=0.3669\times 3.7\times {10}^{10}dps$; initial activity of $X^{A_2}=0.6331\times 3.7\times {10}^{10}dps$

Let activity of $X^{A_1}$ and $X^{A_2}$ are a and b curie, respectively, at t = 0.
$\therefore$ a + b = 1 curie ...... (i)
For $X^{A_1}:t=\frac{2.303}{K}log\frac{N_0}{N}$
$=\frac{2.303}{K}log\frac{r_0}{r}$
$20=\frac{2.303\times 25}{0.693}log\frac{a}{r_1}$
$r_1=0.3716a$
For $X^{A_2}:-t=\frac{2.303}{K}log\frac{N_0}{N}$
$=\frac{2.303}{K}log\frac{r_0}{r}$
$20=\frac{2.303\times 25}{0.693}log\frac{b}{r_2}$
$r_2=0.5744b$
Given activity after 20 days $=\frac{1}{2}$ curie
$0.3716a+0.5774b=\frac{1}{2}$
0.7432 a + 1.1488 b = 1 .... (ii)
By solving equations (i) and (ii), we get
$a=0.3669c=0.3669\times 3.7\times {10}^{10}$ dps
$b=0.6331c=0.6331\times 3.7\times {10}^{10}$ dps
Now rate = KN
($\therefore$ a = 0.3669 curies)
For $X^{A_1}:0.3669\times {10}^{10}\times 3.7=\frac{0.693\times N_0^{A_1}}{14\times 24\times 60\times 60}$
For $X^{A_2}:0.6331\times {10}^{10}\times 3.7=\frac{0.693\times N_0^{A_2}}{25\times 24\times 60\times 60}$
$\Rightarrow \frac{N_0^{A_1}}{N_0^{A_2}}=0.3245$