1 ,mg gold undergoes decay with 2.7 days half | vedclass

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Question

(d) $N = {N_0}{\left( {\frac{1}{2}} ight)^n} $

$\Rightarrow N = {N_0}{\left( {\frac{1}{2}} ight)^{t/{T_{1/2}}}}$

$ \Rightarrow N = 1 	imes

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(d) $N = {N_0}{\left( {\frac{1}{2}} ight)^n} $

$\Rightarrow N = {N_0}{\left( {\frac{1}{2}} ight)^{t/{T_{1/2}}}}$

$ \Rightarrow N = 1 	imes {\left( {\frac{1}{2}} ight)^{\frac{{8.1}}{{2.7}}}} = {\left( {\frac{1}{2}} ight)^3} = \frac{1}{8}$

$ \Rightarrow N = \frac{1}{8}mg = 0.125\;mg$

$1 \,mg$ gold undergoes decay with $2.7$ days half-life period, amount left after $8.1$ days is ......... $mg$

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