If 10% of a radioactive material decays in 5 | vedclass

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Question

(b) $N = {N_0}{e^{ - \lambda t}}$

$	herefore \;0.9{N_0} = {N_0}{e^{ - \lambda  	imes 5}} $

$\Rightarrow 5\lambda  = {\log _e}\frac{1

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(b) $N = {N_0}{e^{ - \lambda t}}$

$	herefore \;0.9{N_0} = {N_0}{e^{ - \lambda  	imes 5}} $

$\Rightarrow 5\lambda  = {\log _e}\frac{1}{{0.9}}$    ...  $(i)$

and $x{N_0} = {N_0}{e^{ - \lambda  	imes 20}} $

$\Rightarrow 20\lambda  = {\log _e}\left( {\frac{1}{x}} ight)$      ... $(ii)$

Dividing $(i)$ by $(ii)$, we get

$\frac{1}{4} = \frac{{{{\log }_e}(1/0.9)}}{{{{\log }_e}(1/x)}} $

$= \frac{{{{\log }_{10}}(1/0.9)}}{{{{\log }_{10}}(1/x)}} = \frac{{{{\log }_{10}}0.9}}{{{{\log }_{10}}x}}$

$ \Rightarrow {\log _{10}}x = 4{\log _{10}}0.9 $

$\Rightarrow x = 0.658 = 65.8\% $

If $10\%$ of a radioactive material decays in $5$ days, then the amount of original material left after $20$ days is approximately ...............$\%$

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