Consider a system of three molecules A, B, and C. Suppose that three units of energy can be distributed over the three atoms. Each atom can have no energy, one unit of energy, two units of energy, or all three units of energy. Assume that each of the ten possible arrangements of the three units of energy is equally probable. What is the probability that molecule B has exactly one unit of energy?
Each atom of three, has four states, and each are equal probable.
Possible configurations...
A,B,C
1,1,1
1,2,0
1,0,2
2,1,0
2,0,1
3,0,0
0,3,0
0,0,3
0,1,2
0,2,1
To solve this problem, we need to consider all the possible arrangements of energy units among the three molecules A, B, and C. Let's list all the possible outcomes first.
We have three units of energy, and each molecule can have a range of 0 to 3 units. Here are all the possible arrangements:
1. A = 0, B = 0, C = 3
2. A = 0, B = 1, C = 2
3. A = 0, B = 2, C = 1
4. A = 0, B = 3, C = 0
5. A = 1, B = 0, C = 2
6. A = 1, B = 1, C = 1
7. A = 1, B = 2, C = 0
8. A = 2, B = 0, C = 1
9. A = 2, B = 1, C = 0
10. A = 3, B = 0, C = 0
We have a total of 10 equally probable outcomes. Out of these 10 possibilities, we need to find the arrangements where molecule B has exactly one unit of energy.
From the above list, we can see that there are two arrangements where molecule B has exactly one unit of energy:
1. A = 0, B = 1, C = 2
2. A = 1, B = 1, C = 1
So, the probability that molecule B has exactly one unit of energy is 2 out of 10, which can be simplified to 1/5.
Therefore, the probability is 1/5.