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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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The activity of a radiooctine sample is measured as 4750 counts/min at $\;t=0\;$ and as 2700 counts/min at $\;t=5$ minutes . Calculate Lay life of sample

$(a)\;7.2 min\qquad(b)\;6.13 min\qquad(c)\;5.24 min\qquad(d)\;8.3 min$

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Answer : 6.13 min
Explanation :
According to decay equation
$\lambda t =ln(\large\frac{N_{0}}{N}) \qquad or \qquad \lambda=\large\frac{1}{t}\;ln(\large\frac{N_{0}}{N})$
$or\; \lambda=\large\frac{1}{t} \times2.303 \qquad ln(\large\frac{N_{0}}{N})-----(1)$
$Now \; \large\frac{d N_{0}}{dt}=\lambda N_{0}\; \quad and \; \large\frac{dN}{dt}=\lambda N$
Hence $\; \large\frac{N_{0}}{N}=\large\frac{\large\frac{dN_{0}}{dt}}{\large\frac{dN}{dt}}=\large\frac{4750}{2700}=1.76-----(2)$
From eq (1) & (2)
$\lambda=(\large\frac{1}{t})\times2.303\times ln(1.76)$
$=(\large\frac{1}{5} minutes)\times2.303\times ln(1.76)$
$=0.113 /minute$
$T=\large\frac{0.693}{\lambda}=\large\frac{0.693}{0.113}=6.13minutes$
answered Feb 24, 2014 by yamini.v
edited Feb 24, 2014 by yamini.v
 

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