Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Atoms

Radio isotopes of phosphorous $\;P^{32}\;$ and $\;P^{35}\;$ are mixed in ratio of 2:1 of atoms .The activity of samples 2 $\;\mu Ci\;$.Find activity of sample after 30 days. $\;t_{\large\frac{1}{2}}\;$ of $\;P^{32}=14\;$days and $\;t_{\large\frac{1}{2}}\;$ of $\;P^{33}\;$ is 25 days.

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

1 Answer

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
Similarly for final activity (Activity after time 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
On solving (1) & (2) we get $\;A_{20}=\large\frac{7}{16}\;$& $\;A_{10}=\large\frac{25}{16}$
$A_{t}=A_{10}e^{-\lambda_{1} t}+A_{20} e^{-\lambda_{2}t}$
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 $=\large\frac{25}{16}\times0.2265+\large\frac{7}{16}\times0.4353$
$=0.54 Ci$
answered Feb 24, 2014 by yamini.v

Related questions