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

In an experiment, the activity of 1.2 mg of radioactive potassium chloride (chloride of isotope of K-40) was found to be `170s^-1`: Taking molar mass to be `0.075 kg "mole"^-1` find the number of `K-40` atoms in the same and hence find the half life of `K-40`. Avogadro's number `=6.0xx10^(23)"mole"^-1`

Text Solution

Verified by Experts

Here, m=1.2mg=`1.2xx10^-3g`
`A=170s^-1, M=0.075kg=75g, T=?`
Number of molecules present in the sample
`N=m/MxxN_A=(1.2xx10^-3)/75xx6.0xx10^(23)`
`=9.6xx10^(18)`
As `A=lambdaN=0.693/TxxN`
`T=(0.693N)/A=(0.693xx9.6xx10^(18))/170`
`=3.91xx10^(16)sec`
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