Six distinguishable particles are distributed over three non-degenerat...
Since, the levels are non-degenerate there is only one state associated with each energy.
Let the number of particles is 3 energy state be N
1 , N
2 and N
3 respectively.
where N
1 + N
2 + N
3 = 6
As the particles are distinguishable, the number of microstates i.e. the number of ways of choosing N
1 , N
2 and N
3 particles from 6 particles is
W → thermodynamic probability. It is maximum when N
1 !N
2 !N
3 ! is a minimum
where N
1 = N
2 = N
3 = 2
The corresponding total energy of distribution is
= 6E
The correct answer is: 6
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Six distinguishable particles are distributed over three non-degenerat...
Introduction:
In this problem, we are given six distinguishable particles that are distributed over three non-degenerate energy levels. The energy levels are 0, E, and 2E. We need to determine the total energy of the distribution for which the probability is a maximum.
Understanding the problem:
To understand the problem, let's consider the possible ways in which the particles can be distributed among the energy levels. Since the particles are distinguishable, we need to consider the permutations of the particles.
Possible distributions:
Let's consider the possible distributions of the particles among the energy levels:
1. All particles in the 0 energy level: In this case, there is only one possible distribution.
2. One particle in the E energy level and the remaining five particles in the 0 energy level: In this case, we have six choices for the particle in the E energy level, and for each choice, we have five choices for the remaining particles in the 0 energy level. So, the total number of distributions is 6 * 5 = 30.
3. Two particles in the E energy level and the remaining four particles in the 0 energy level: In this case, we have six choices for the first particle in the E energy level, and for each choice, we have five choices for the second particle in the E energy level. Similarly, we have four choices for the remaining particles in the 0 energy level. So, the total number of distributions is 6 * 5 * 4 = 120.
4. One particle in the 2E energy level and the remaining five particles in the 0 energy level: Similar to case 2, we have six choices for the particle in the 2E energy level, and for each choice, we have five choices for the remaining particles in the 0 energy level. So, the total number of distributions is 6 * 5 = 30.
5. One particle in the E energy level, one particle in the 2E energy level, and the remaining four particles in the 0 energy level: In this case, we have six choices for the particle in the E energy level, and for each choice, we have five choices for the particle in the 2E energy level. Similarly, we have four choices for the remaining particles in the 0 energy level. So, the total number of distributions is 6 * 5 * 4 = 120.
6. Two particles in the 2E energy level and the remaining four particles in the 0 energy level: Similar to case 3, we have six choices for the first particle in the 2E energy level, and for each choice, we have five choices for the second particle in the 2E energy level. Similarly, we have four choices for the remaining particles in the 0 energy level. So, the total number of distributions is 6 * 5 * 4 = 120.
Determining the maximum probability:
To determine the distribution for which the probability is a maximum, we need to consider the distribution with the highest number of possible arrangements. From the above analysis, we can see that cases 3, 5, and 6 have the highest number of distributions (120). Therefore, the total energy of the distribution for which the probability is a maximum is 2E.
Conclusion: