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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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The disintegration rate of a radioactive sample at a certain instant is 4750 dpm . Five minute later rate becomes 2700 disintegration per minute . Then the decay constant is

$(a)\;0.26 per \;min\qquad(b)\;0.113 per\; min\qquad(c)\;0.35 per\; min\qquad(d)\;0.05 per\; min$

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1 Answer

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Answer : $\;0.113\; per\; min$
The rate of disintegration R is given by $\;R=R_{0} e^{-\lambda t}$, where $\;R_{0}\;$ is initial rate at t=0
This equation gives $\;\large\frac{R_{0}}{R}$$=e^{\lambda t} \rightarrow \ln \large\frac{R_{0}}{R}$$=\lambda t$
$\Rightarrow \lambda=\large\frac{1}{t}\;$$\ln\large\frac{R_{0}}{R}$$=\large\frac{2.303}{t}\;$$\log_{10} \large\frac{R_{0}}{R}$
$R_{0}=4750 \qquad R=2700 \qquad t=5 min$
$\lambda=\large\frac{2.303}{5}$$\;\log_{10} \large\frac{4750}{2700}$$=\large\frac{2.303}{5} $$\log_{10} (1.76)$
$\Rightarrow \;\lambda=\large\frac{2.303}{5}$$\times0.2455=0.1131 \;per\;min\;.$
answered Feb 26, 2014 by yamini.v
edited Aug 12, 2014 by balaji.thirumalai
 

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