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The half life of a radioactive substance is `5xx10^(3)`yrs. In how many years will its activity decay to 0.2 times its initial activity? Take `log_(10) 5=0.6990`. |
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Answer» `T= 5000` years, `(N)/(N_(0))=0.2=(2)/(10)=(1)/(5)` `lambda=(0.693)/(T)=(0.693)/(5000)` `(N)/(N_(0))=e^(-lambdat)` `(1)/(5)=(1)/(e^(lambdat))rArr5=e^(lambdat)` `log_(e )^(5)=lambdat` `t=(2.303xx0.6990xx5000)/(0.693)` `t=11614.6` years `=1.1615xx10^(4)` years |
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