If 20, gm of a radioactive substance due to radioactive decay reduces to 10 ,gm in 4 ,minutes, then

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13.Nuclei

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If $20\, gm$ of a radioactive substance due to radioactive decay reduces to $10 \,gm$ in $4 \,minutes,$ then in what time $80\, gm $ of the same substance will reduce to $10 \,gm$

A

In $8 \,minutes$

B

In $12\, minutes$

C

In $16\, minutes$

D

In $20\, minutes$

Solution

(b) $20\, gm$ substance reduces to $10\, gm$ ($i.e.$ becomes half in $4\, min.)$

So ${T_{1/2}} = 4\,\min $.

Again $M = {M_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}}$

==> $10 = 80\,{\left( {\frac{1}{2}} \right)^{t/4}}$

==>$\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = {\left( {\frac{1}{2}} \right)^{t/4}}$

==>$ t = 12\, min.$

Standard 12

Physics

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