A radio-active material is reduced to 1 / 8 o | vedclass

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Question

$N = N _0\left(\frac{1}{2}ight)^{ n }$

$\frac{ N _0}{8}= N _0\left(\frac{1}{2}ight)^{ n }$

$n =3$

$3 	ext { half lives }=3 	ext { days

Vedclass · Accepted Answer

$256$

Vedclass · Answer

$N = N _0\left(\frac{1}{2}ight)^{ n }$

$\frac{ N _0}{8}= N _0\left(\frac{1}{2}ight)^{ n }$

$n =3$

$3 	ext { half lives }=3 	ext { days }$

$1 	ext { half life }=1 	ext { day }$

$5 	ext { days }=5 	ext { half life }$

$N = N _0\left(\frac{1}{2}ight)^{ n }$

$8 	imes 10^{-3}= N _0\left(\frac{1}{2}ight)^5$

$N _0=256 	imes 10^{-3}\,kg$

$N _0=256\,g$

A radio-active material is reduced to $1 / 8$ of its original amount in $3$ days. If $8 \times 10^{-3}\,kg$ of the material is left after $5$ days. The initial amount of the material is $.......\,g$

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