Question

⁸¹Sr decays by positron emission with a half-life of 22.2 minutes. Starting with 885g of Sr-81, how much remains after 75.0 minutes?

Answer 1

Given :Half life : 22.2 min

Initial amount of Sr-81= [ Sr -81 ] _{t = 0} = 885 g  

Amount of Sr -81 left after time 75.0 min = [ Sr -81 ] _{t =75.0 min} = ?

Time of disintegration =75.0 min

We have equation: t =2.303 /    log (N ₀ / N _t) -------------------> (1)

Where, is a decay constant, N ₀ is the initial amount of radioactive element, N _t is the amount of radioactive element after time t.

t =2.303 /    log [ Sr -81 ] _{t = 0} /  [ Sr -81 ] _{t =75.0 min}

We have relation , decay constant = 0.693 / t _1/2

decay constant = 0.693 / 22.2 min = 0.03122 min ^-1

75.0 min =  2.303 / 0.03122 min ^-1 log 885 g /  [ Sr -81 ] _{t =75.0 min}

log 885 g /  [ Sr -81 ] _{t =75.0 min} = 75.0 min 0.03122 min ^-1 / 2.303

log 885 g /  [ Sr -81 ] _{t =75.0 min} = 1.017

885 g /  [ Sr -81 ] _{t =75.0 min} = 10 ^1.017

885 g /  [ Sr -81 ] _{t =75.0 min} = 10.40

  [ Sr -81 ] _{t =75.0 min} = 885 g / 10.40 = 85.1 g

ANSWER : Amount of Sr-81 left after 75.0 min = 85.1 g