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The counting rate observed from a radioactive source at t = 0 seconds was 1600 counts / sec and at t = 8 sec it was 100 counts /sec . The counting rate obserbed as count per sec at t = 6 sec will be

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Solution :`lambda XX 8 = 2.303 log ((600)/(100)) , lambda = (2.303)/(8) xx 4 log (2) = (2.303 xx 4 xx 0.3010)/(8) = (0.693)/(2)`
`t_(1//2) = (0.693)/(lambda) = (0.693)/(0.693//2) = 2` sec , `1600 overset(2) (to) 800 overset(2)(to) 400 overset(2) (to) 200 , T = 6`


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