By ⁺ -emission, ⁵⁷ Co decays into ⁵⁷ Fe . The produced ⁵⁷ Fe is initially in an excited state and transitions to the ground state by emitting -rays. The half-life for the ⁺ -decay is 270 days , while the half-life for the -emission is 10⁻⁸ s . If a ⁵⁷ Co sample currently emits 5.0 10⁹ gamma rays per second, what time will elapse until the gamma ray emission rate drops to 2.5 10⁹ per second?

Step-by-step solution

The decay sequence is ⁵⁷ Co ^+ ⁵⁷ Fe ^ * ⁵⁷ Fe . Since the half-life of ⁵⁷ Co ( 270 days ) is much greater than the half-life of the excited ⁵⁷ Fe ^ * state ( 10⁻⁸ s ), secular equilibrium is established almost instantaneously. In secular equilibrium, the rate of formation of the daughter nucleus is equal to its rate of decay. Thus, the activity of ⁵⁷ Fe ^ * ( -emission rate) is equal to the activity of ⁵⁷ Co ( ⁺ -emission rate). The -emission rate will therefore decay with the half-life of the parent ⁵⁷ Co nucleus