What is the disintegration constant of a radioactive element, if the number of its atoms diminishes by 18 % in 24 h ? Where ln 50 41 = 0 . 198

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The percentage of element of left behind after time t,

$\frac{N}{N_{0}} = (100 - 18) % = 82 %$

$∵ \frac{N}{N_{0}} = e^{- λ t}$

$\Rightarrow \frac{82}{100} = e^{- λ (24 \times 60 \times 60)}$

$\Rightarrow λ \times 24 \times 60 \times 60 = \ln \frac{100}{82}$

$\Rightarrow λ = \frac{0.198}{24 \times 60 \times 60} = 22 \times 10^{- 6} s^{- 1}$

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