Asked by Beth
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
I keep getting the wrong answer and use the equation H=kNo
Answers
Answered by
Elena
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
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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