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

## Answer This Question

## Related Questions

- chemistry - Determine the activity of 1.11 g of 88Ra226 (a) in disintegrations ...
- Chemistry - The cloth shroud from around a mummy is found to have a carbon-14 ...
- chem - I am not sure if there is an equation for this problem or wut?? A 73-kg ...
- college geology help please - I was the same person from yesterday. I need help ...
- college geology dealing with carbon 14 - ok in my professors notes he calculates...
- college Geol 101 - Suppose you have a sample of clamshell at a Paleoindian site ...
- chemistry - I have: the half life of U-238 is 4.5 x 10^9 yr. A sample of rock of...
- ChemB - The activity of carbon 14 in living tissue is 15.3 disintegrations per ...
- Chemistry - I don't know if I should be posting this on a new session or just ...
- chem - If a radioactive sample has an activity of 8.33 Bq, how many ...

More Related Questions