1.

A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration into becquerel. (one curie = 3.7xx10^(10) Bq)

Answer»

Solution :Cobalt specimen emits induced RADIATION = 75.6 millicurie per second ( `"1 curie" = 3.7 x 10^(10)` BQ)
So 75.6 millicurie = `75.6 xx 10^(-3) xx 1 "curie"`
`75.6 xx 10^(-3) xx 3.7 xx 10^(10)` Bq
`= 279.72 x 10^(7)`
`= 2.7972 xx 10^(9) "Bq"`
75.6 millicurie per second is equivalent to `2.7972 xx 10^(9) "Bq".`


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