1.

Percentage of a radioactive element decayed after 20 s when half-life is 4 s

Answer»

92.25
96.87
50
75

Solution :`n = (20)/(4) = 5, (N_(t))/(N_(0)) = ((1)/(2))^(5) = (1)/(32), :.` decayed
`= (1-(1)/(32)) XX 100 = (31)/(32) xx 100 = 96.87`


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