Please help me with this problem.
The standard source for radiation therapy was radioactive 60Co which decays with a halflife of 5.27y, into an excited nuclear state of 60Ni. That nickel isotope then immediately emits two gamma ray photos wach with energy of 1.2 MeV. How many radioactive 60Co nuckei are present in a 6000 Ci source?
Cheers
The decay process is explained better at
http://en.wikipedia.org/wiki/Cobalt-60
Each Co-60 nucleus results in one beta electron and two gamma photons. That is three nearly simultaneous radioactive emission processes during the each transmutation, with a 5.27 year half life.
One gram of Co60 has a radioactivity level of 50 Ci. One should be able to derive that from the half life and the three radioactive decay steps per nucleus. 6000 Ci would therefore require 120 g of Co-60. That equals 2.00 moles, so there are 12*10^23 Co-60 nuclei in that amount of Curies.
To find out the number of radioactive 60Co nuclei present in a 6000 Ci source, we need to utilize the concept of radioactive decay and the equation for the number of radioactive nuclei remaining over time.
The decay of a radioactive substance can be described by the equation:
N(t) = N₀ * (1/2)^(t/τ)
Where:
N(t) is the number of radioactive nuclei remaining at time t
N₀ is the initial number of radioactive nuclei
τ is the half-life of the substance
t is the elapsed time
Here, in the given problem, we are provided that the half-life of 60Co is 5.27 years. We need to find the initial number of 60Co nuclei for a 6000 Ci source.
To convert from Ci (Curie) to the number of radioactive nuclei, we can make use of Avogadro's number, which states that 1 mole of a substance contains 6.022 × 10^23 particles (atoms, molecules, or nuclei).
First, we convert from Ci to Bq (Becquerel). The conversion factor is 1 Ci = 3.7 × 10^10 Bq.
So, for a 6000 Ci source:
Activity (Bq) = 6000 Ci * 3.7 × 10^10 Bq/Ci
Now, we can use the definition of activity (A) as the rate of decay, given by:
A = λ * N
Where:
A is the activity in Bq
λ is the decay constant (λ = ln(2)/τ)
N is the number of radioactive nuclei
Rearranging the equation, we solve for N:
N = A / λ
In this case, λ = ln(2) / τ.
Now, we can substitute the values to find N. The decay constant, λ, is calculated as:
λ = ln(2) / 5.27 years
N = (Activity (Bq)) / λ
Let's calculate N.
Activity (Bq) = 6000 Ci * 3.7 × 10^10 Bq/Ci
λ = ln(2) / 5.27 years
N = (Activity (Bq)) / λ
N = [(6000 Ci * 3.7 × 10^10 Bq/Ci)] / [ln(2) / 5.27 years]
By performing the calculations, you can find the number of radioactive 60Co nuclei present in the 6000 Ci source.