1.

The half-lifeperiod of ………… is 60days.The radioactivity after 180 days will be

Answer»

0.25
0.125
`33.3%`
`3.0%`

Solution :RADIOACTIVITY after time,t
`(N_t) = (N_0)/((2)^(n)) and n = t/(t^(1//2))`
So, `n = 180/60 = 3, N_t = (N_0)/(8) = 0.125`
Hence, radioactivity after 180 days = 12.5%.


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