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In an experiment, the activity of 1.2 mi...

In an experiment, the activity of 1.2 milligrams of radioactive potassium chloride `K^(40)`Cl was found to be 17.5%. Taking molar mass of `K^(40)` CI to be 0.075 kg/mol. Find the number of atoms in the sample and the half-life of `K^(40)`. Take avogadro number `6.0 xx 10^(23) mol^(-1)` .

Text Solution

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`0.075 ` kg of the material containing `6xx10^(23)` atoms of `K^(40)` . Therefore the number of atoms in 1.2 mg =` 1.2 xx10^(-6)` kg of the material is
`N = (6xx 10^(23)xx 1.2 xx 10^(-6))/(0.075) = 96 xx 10^(17) `
The activity R is given by
`implies R = (-dN)/(dt) = lambda N = (0.6931)/(T )N`
where T is the half - life of `K^(40)`
Here R = 170 `s^(-1)`
`T = (0.6931 N )/(R ) = (0.6931 xx 96xx10^(17))/(170)`
`= 0.39 xx 10^(17) ` s
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