After 3 hours, only 0.25 ,mg of a pure radioa | vedclass

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Question

(b)

Initial mass $=2 \,mg$

Final mass $=0.25 \,mg$

$	herefore \frac{N}{N_0}=\left(\frac{1}{2}ight)^n$

$\frac{0.25}{2}=\left(\frac{1}{2}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(b)

Initial mass $=2 \,mg$

Final mass $=0.25 \,mg$

$	herefore \frac{N}{N_0}=\left(\frac{1}{2}ight)^n$

$\frac{0.25}{2}=\left(\frac{1}{2}ight)^n$

$\frac{1}{8}=\left(\frac{1}{2}ight)^n$

$	herefore n=3$ and time is $3$ hours

Half life is therefore $1$ hour

After $3$ hours, only $0.25 \,mg$ of a pure radioactive material is left. If initial mass was $2 \,mg$ then the half life of the substance is ...... $hr$

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