Question

The half-life of `""_82^210Pb` is 22.3 y. How long will it take for its activity 0 30% of the initial activity?

Answer 1

T = 22.3 y, A = 0.3 A₀

By the radioactive decay law, N = N₀e^−λt

∴ λN = λN₀e^−λt

∴ A = A₀e^−λt

where A₀ = λN₀ is the initial activity and A = λN is the activity at time t.

∴ λt = log_e `A_0/A = 2.303 log_10  (A_0)/A`

∴ t = `2.303/λ log_10  (A_0)/A`

= `(2.303 T)/0.693 log_10  (A_0)/A`      (∵ λ = `0.693/T`)

= `(2.303 xx 22.3)/0.693 log_10 (10/3)`

= `(2.303 xx 22.3)/0.693 (log_10 10 - log_10 3)`

= `(2.303 xx 22.3)/0.693 (1.0000 - 0.4771)`

log 2.303     =   0.3623
log 22.3      = + 1.3483
log 0.5229  = + `underline(overline(1).7184)`
                          1.4290
log 0.693    = `underline(- overline(1).8407`
                          1.5883
antilog 1.5883 = 38.76

= `(2.303 xx 22.3)/0.693 xx 0.5229`

= 38.76 y

It will take 38.76 y for the activity of `""_82^210 Pb` to reduce to 30% of the initial activity.