1.

1.0 g of a radioactive isotope was found to reduce to 125 mg after 24 hours. The half-life of the isotope is

Answer»

8 HOURS
24 hours
6 hours
4 hours

Solution :`N = [(1)/(2)]^(n) xx N_(0) = 125mg = ((1)/(2))^(n) xx 1000 mg`
`((1)/(2))^(n) = (125)/(1000) = (1)/(8)`
`((1)/(2))^(n) = ((1)/(2))^(3),n = 3`, so number, of `t_(1//2) = 3`
TOTAL time = 24 hours, Half-life time = `(24)/(3)` = 8 hours.


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