Question

A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 10¹⁰ Bq)

Answer 1

Cobalt specimen emits induced radiation = 75.6 millicurie per second
(1 curie = 3.7 × 10¹⁰ Bq)
So 75.6 millicurie = 75.6 × 10³ × 1 curie
= 75.6 × 10^-3 × 3.7 × 10¹⁰ Bq
= 279.72 × 10⁷
= 2.7972 × 10⁹ Bq
75.6 millicurie per second is equivalent to 2.7972 × 10⁹ Bq.