1.

After 24 hrs, only 0.125 g out of the initial quantity of 1 g of a radioactive isotope remains behind. What is its half-life period?

Answer»

Solution :The DATA given can be WRITTEN as under:
`a=1g, a-x=0.125g, t=24` hours
Substituting the VALUES in the first order equation
`K=(2.303)/(t)"LOG"(a)/((a-x))`, we have
`k=(2.303)/(24h)xx"log"(1)/(0.125)`
We have, `k=(2.303)/(24h)xx log 8=0.0866 h^(-1)`
`t_(1//2)=(0.693)/(k)=(0.693)/(0.0866)"hours"=8` hours


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