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Disintegration rate of a sample is 10^{10} per hour at 20 hours from the start. It reduces to 6.3 x 10^{9} per hour after 30 hours. Calculate its half-life and the initial number of radioactive atoms in the sample.

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#### Solution

**Data:** A(t_{1}) = 10^{10} per hour, where t_{1} = 20 h,

A(t_{1}) = 6.3 × 10^{10} per hour, where t_{2} = 30 h

A(t) = A_{0}e^{-λt } ∴ A(t_{1}) = `"A"_0"e"^(-λ"t"_1)`^{ }and

A(t_{2}) = `"A"_0"e"^(-λ"t"_2)`

∴ `("A"("t"_1))/("A"("t"_2)) = ("e"^(-lambda"t"_1)/"e"^(-lambda"t"_2)) = "e"^(lambda("t"_2 - "t"_1))`

∴ `10^10/(6.3 xx 10^9) = "e"^(lambda(30 - 20)) = "e"^(10lambda)`

∴ 1.587 = e^{10λ}

∴ 10λ = 2.303 log_{10}(1.587)

∴ λ = (0.2303)(0.2007) = 0.04622 per hour

The half life of the material, T_{1/2} = `0.693/lambda = 0.693/0.04622`

= 14.99 hours

Now, `"A"_0 = "A"("t"_1)"e"^(lambda"t"_1) = 10^10"e"^((0.04622)(20))`

= `10^10 "e"^0.9244`

Let x = `"e"^0.9244`

∴ 2.303 log_{10}x = 0.9244

∴ log_{10}x = `0.9244/2.303 = 0.4014`

∴ x = antilog 0.4014 = 2.52

∴ A_{0} = 2.52 x 10^{10} per hour

Now A_{0} = N_{0}λ

∴ `"N"_0 = "A"_0/lambda = (2.52 xx 10^10)/0.04622`

= 5.452 × 10^{11}

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