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Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy. Repeat the calculation given in the text for the relative probability of being in a first excited state, taking spin degeneracy into account. Show that the results are unaffected.

Short Answer

Expert verified

Therefore,

PE2PE1=4e-E2-E1/kT

Step by step solution

01

Given information

Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy.

02

Explanation

With n = 2, if E1 is the ground state energy and E2 is the first excited state energy, the probability is:

P(s)=1Ze-E/kT

There are four states in E2 and one state in E1, thus their probabilities are P(s2) multiplied by eight and P(s1) multiplied by two when we add the spin. Thus, the probability of E2 equals Ps2multiplied by eight and the probability of E1 equals Ps1multiplied by two when we include the spin.

PE2=8Ze-E2/kTPE1=2Ze-E1/kT

The ratio of these probabilities is:

role="math" localid="1647300792412" PE2PE1=8e-E2/kT2e-E1/kTPE2PE1=4e-E2-E1/kT

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Most popular questions from this chapter

A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the "flexing' mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of 4.8 x 1013Hz. As for any quantum harmonic oscillator, the energy levels are 12hf,32hf,52hf, and so on. None of these levels are degenerate.

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