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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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A small quantity of solution containing $\;Na^{24}\;$ radio nuclide (Half life 15 hours ) of activity 1 micronuclids is injected into blood of a person A sample of blood of volume $\;1 cm^{3}\;$ taken after 5 hours sheemes on activity of 296 disintegration per minute. The total volume of blood in body of person is (Assume that radioactive solution mixes uniformly in blood of person . $\;(1 Ci =3.7\times10^{10} dps)$)

$(a)\;6 l\qquad(b)\;8 l\qquad(c)\;2 l\qquad(d)\;1 l$

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Answer : (a) 6 l
Explanation :
The activity varries according to equation $\;A (t) = A_{0} \;e^{-\lambda t}$
or $\;A(t)=A_{0} A_{0} \;(\large\frac{1}{2})^{\large\frac{t}{T_{\large\frac{1}{2}}}}$
$A (t=5 hr)=296 \;disintegrations/min$
$\large\frac{296}{60}=A_{0}\; (\large\frac{1}{2})^{\large\frac{5}{15}}$
$A_{0}=\large\frac{296}{60} \times2^{\large\frac{1}{3}}=6.22\;disintegration/sec $
= the initial activity of 1cc of blood .
If total volume of blood is $V_{cc}.$
$6.22 V=3.7\times10^{6} \times10^{-6}=6 l\;.$
answered Feb 26, 2014 by yamini.v
 

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