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Q. The activity of a radioactive sample is measured as $9750$ count/min at $t=0$ and 975 count/ $\min$ at $t=5 \,min$. The decay constant is nearly

Bihar CECEBihar CECE 2010Nuclei

Solution:

Activity, $A=A_{0} e^{-\lambda t}$
$\therefore 975=9750 e^{-\lambda \times 5}$
$\frac{1}{10}=e^{-\lambda \times 5}$
Taking logarithm on both sides, we have
$-\log _{e} 10=-5 \,\lambda$
$-2.3026=-5\, \lambda$
$\lambda=\frac{2.3026}{5}=0.461 \,min ^{-1}$