Posted by **Beth** on Sunday, July 29, 2012 at 12:18am.

Determine the activity of 1.11 g of 88Ra226 (a) in disintegrations per second (Bq) (b) and in Curies (Ci). The half-life is 1600 yr (1 yr = 3.156 x 10^7 s). 1 Ci is equal to 3.7 x 10^10 Bq.

I keep getting the wrong answer and use the equation H=kNo

- Physics -
**Elena**, Sunday, July 29, 2012 at 6:53am
A=λ•N=(ln2/T)•N

The amount of Radium atoms is

N=m•N(A)/μ,

where

N(A) = 6.022•10^23 mol^-1 is the Avogadro Number,

μ = 0.226 kg/mol is the molar mass of Radium

A=λ•N=(ln2/T) •m•N(A)/μ =

=(0.693•1.11•10^-3•6.022•10^23)/(1600•3.156•10^7•0.226)=4.06•10^10 Bq=1.1 Ci

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