$(a)\;6\times 10^{-5}\qquad(b)\;6\times 10^5\qquad(c)\;3\times 10^5\qquad(d)\;3\times 10^{-5}$

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$[A]^{-n+1}=[A]^{-2}$

$n=3$

$(3-1)k=20\times 10^{-6}$

$K=10\times 10^{-6}$

$K=10^{-5}$

$t_{\large\frac{1}{2}}=\frac{1}{K(n-1)}\bigg[\frac{2^{n-1}-1}{[A_o]^{1-n}}\bigg]$

$\Rightarrow \large\frac{1}{10^{-5}(2)}\bigg[\frac{2^2-1}{(2)^{1-3}}\bigg]$

$\Rightarrow \large\frac{3\times 4\times 10^5}{2}$

$\Rightarrow 6\times 10^5$

Hence (b) is the correct answer.

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