# Branching fraction

• I
leoneul
I’ve had huge difficulty understanding/interpreting the concept of branching fraction. So correct me if I’m wrong please:
Let’s take the decay of Ra-226 with half-life of 1602 years as an example. It decays through alpha 1 chanel to the excited state of Rn-222 with E=0,187 Mev ( branching fraction 5,4%) and through alpha 2 decay chanel to the ground state of Rn-222 ( Branching fraction 94,6%).

Now my understanding/interpretation :
If we have for example 10^6 Ra-226 nuclides at t=0 , then after 1602 years have passed , 0.054 * ( 10^6 / 2 )=27000 of alpha1 particles and (10^6/2)- 27000=527000 alpah 2 particles will have emitted.

Homework Helper
Hello leoneul,

Idea's good, math is not : (10^6/2)- 27000=473000

leoneul
leoneul
Hello leoneul,

Idea's good, math is not : (10^6/2)- 27000=473000
hehe thanks alot! This simple concept has been causing so much trouble bcs some sources refers to the branching ratio as “ probability” which confused me

For anyone atom the branching ratio reflects the probabilities for the channels. Don't overthink!

leoneul
leoneul
For anyone atom the branching ratio reflects the probabilities for the channels. Don't overthink!
So for my example the probability that a certain SINGLE Ra-226 nucleus decays through alpha 1 channel is 5.4% but if we have an ENSEMBLE of Ra-226, then 5.4 % will decay through chanel 1. Correct?

Again, ideas correct. This time the wording could be a bit sharper by expressing it as a ratio $${\alpha_1\ {\rm decays} \over {\rm total \ decays}} \times 100 \;\%$$ however, litterally taken what you write is correct (but you have to wait infinitely long ...)
Again, ideas correct. This time the wording could be a bit sharper by expressing it as a ratio $${\alpha_1\ {\rm decays} \over {\rm total \ decays}} \times 100 \;\%$$ however, litterally taken what you write is correct (but you have to wait infinitely long ...)