A 4.00x10^-2 mol sample of a radioactive isotope has an activity of 56.5 Ci. What is the decay constant of the isotope (in/s) ?

I've worked a different example in my textbook but the problem I am having is the example I learned from the Ci was in the form of 1.7x10^-5 therefore I dont know how to convert the 56.5 to the appropriate figure to complete the problem.

So far all I have is 56.5x (3.7x10^10/1.0 Ci) = 2.1x10^12 , Is this how to get the proper figure for Ci from 56.5? I'm not sure how to go about the rest of the problem

## The rate of decay is often

The rate of decay is often referred to as the activity of the isotope and is often measured in Curies (Ci), one curie = 3.700 x 10

^{10}atoms that decay/second .I will simply consider this one a first order reaction and the rate of radioactive decay with first order is :

r = k[N]

^{1}N is the amount of radioisotope

k is the first order rate constant for the isotope

First, we need to convert the 4.00x10^-2 mol sample of a radioactive isotope into number of atoms of cobalt-60 and to convert the activity into numbers of atoms that decay per second.

1 mol = 6.022 x 10

^{23}atoms of sample givenso, in 4.00x10^-2 mol sample we will have

4.00x10^-2 mol x 6.022 x 10

^{23}atoms = 24 x 10^{21}atomsone curie = 3.700 x 10

^{10}atoms and we have 56.5 Ci given in the problem so the rate is56.5x (3.7x10^10 atoms / s/ Ci) = 2.1x10^12 atoms per decay

Second, we can then use the first-order rate equation to find the rate constant, k

r = k[N]

^{1}2.1x10^12 = k [24 x 10

^{21}atoms]k = ?

It should be prety clear t you now ...