Physical Chemistry , 1st ed.

cn 2  cn (11.15) An equation that relates sequential coefficients like this is called a recursion re- lation.It allows one to ...

index nis the quantum number,and it can have any integer value from 0 to in- finity. (As we will see from the form of the wavefu ...

Solution The difference in energy of the adjacent states is the same and equals h ,or E(6.626 10 ^34 J s)(1.800 1013 s^1 ) ...

To normalize, the wavefunction  0 must be multiplied by some constant N such that N^2    (c 0 ex (^2) /2 )*(c 0 ex (^2 ...

before quantum mechanics was developed. The polynomial parts of the har- monic oscillator wavefunctions are called Hermite polyn ...

By convention, only the positive square root is used. The 2 in the expres- sion above is usually converted into the fourth roo ...

The Hermite polynomial H 3 () contains only odd powers ofx, but upon squaring it becomes a polynomial having only even powers o ...

potential energy. That is, wavefunctions are nonzero and therefore the oscillator can exist beyondits classical turning point. T ...

11.5 The Reduced Mass Many harmonic oscillators are not simply a single mass moving back and forth, like a pendulum or an atom a ...

In considering the total energy of this harmonic oscillation, the potential energy is the same as for any other harmonic oscilla ...

Example 11.9 The hydrogen molecule vibrates at a frequency of about 1.32 1014 Hz. Calculate the following: a.The force constant ...

b.The vibrational frequency expected for a hydrogen atom having a mass of 1.674 10 ^27 kg and a vibrational force constant of ...

where pxand pyare the linear momenta in the xand ydirections. At this time, we are ignoring the vector property of the momenta ( ...

This is not a problem (as inspection of the cosine/sine form of the wavefunc- tion shows). They must be continuous and different ...

The two exponential functions cancel each other out, leaving only the infini- tesimal. The normalization is completed: N^2  2  ...

Example 11.11 An electron is traveling in a circle having radius 1.00 Å. Calculate the energy eigenvalues of the first five 2-D ...

These are the correct units for the moment of inertia. Now we can consider the energies of each state. Since m0 for the first s ...

Solution According to equation 11.39, the values of the angular momenta are  2 ,  1 ,0,1, and 2. Notice that although the ...

Planck’s constant hhas the same units, J s, as the angular momentum, kg m^2 /s. This is a different unit from that oflinearmomen ...

The benzene molecule has a diameterof a little over 3 Å across. The “electron ring” is assumed to be slightly less than that, on ...