# What is the decay constant of the isotope (in/s) ?

akirk798
Mon, 07/17/2017 - 12:21

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 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 ...