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

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#### Solution

T = 22.3 y, A = 0.3 A_{0}

By the radioactive decay law, N = N_{0}e^{−λt}

∴ λN = λN_{0}e^{−λt}

∴ A = A_{0}e^{−λt}

where A_{0} = λN_{0} 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.

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