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In the harmonic oscillators eave functions of figure there is variation in wavelength from the middle of the extremes of the classically allowed region, most noticeable in the higher-n functions. Why does it vary as it does?

Short Answer

Expert verified

Answer:

There is a variation in wavelength from the middle to the extremes because of variation in kinetic energy and hence variation in momentum.

Step by step solution

01

Wavelength

The wavelength gets longer near the extreme edges because the kinetic energy is low there.

As the Kinetic Energy is low, so the momentum is small. Because, we know that Kinetic energy is directly proportional to momentum by the following formula

p=2m(KE)

02

Variations in Wavelengths

And we know that momentum is inversely proportional to wavelength by the formula

ฮป=hp

So, a low kinetic energy corresponds to a longer wavelength.

Hence, Variation of wavelength from middle to extremes is seen because of variation in kinetic energy.

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Most popular questions from this chapter

Obtain expression (5-23) from equation (5-22). Using cosฮธ=cos2(12ฮธ)โˆ’sin2(12ฮธ)andsinฮธ=2sin(12ฮธ)cos(12ฮธ), first convert the argument of the cotangent fromkLto12kL. Next, put the resulting equation in quadratic form, and then factor. Note thatฮฑis positive by definition.

A half-infinite well has an infinitely high wall at the origin and one of finite height U0 at x= L . Like the finite well, the number of allowed states is limited. Assume that it has two states, of energy E1 and E2 , where E2 is not much below U0. Make a sketch of the potential energy, then add plausible sketches of the two allowed wave functions on separate horizontal axes whose heights are E1 and E2 .

Outline a procedure for predicting how the quantum-mechanically allowed energies for a harmonic oscillator should depend on a quantum number. In essence, allowed kinetic energies are the particle-in-a box energies, except the length Lis replaced by the distance between classical tuning points. Expressed in terms of E. Apply this procedure to a potential energy of the form U(x) = - b/x where b is a constant. Assume that at the origin there is an infinitely high wall, making it one turning point, and determine the other numing point in terms of E. For the average potential energy, use its value at half way between the tuning points. Again in terms of E. Find and expression for the allowed energies in terms of m, b, and n. (Although three dimensional, the hydrogen atom potential energy is of this form. and the allowed energy levels depend on a quantum number exactly as this simple model predicts.)

There are mathematical solutions to the Schrรถdinger equation for the finite well for any energy, and in fact. They can be made smooth everywhere. Guided by A Closer Look: Solving the Finite Well. Show this as follows:

(a) Don't throw out any mathematical solutions. That is in region Il (x<0), assume that (Ce+ax+De-ax), and in region III (x>L), assume thatฯˆ(x)=Fe+ax+Ge-ax. Write the smoothness conditions.

(b) In Section 5.6. the smoothness conditions were combined to eliminate A,Band Gin favor of C. In the remaining equation. Ccanceled. leaving an equation involving only kand ฮฑ, solvable for only certain values of E. Why can't this be done here?

(c) Our solution is smooth. What is still wrong with it physically?

(d) Show that

localid="1660137122940" D=12(B-kฮฑA)andF=12e-ฮฑL[(A-Bkฮฑ)sin(kL)+(Akฮฑ+B)cos(kL)]

and that setting these offending coefficients to 0 reproduces quantization condition (5-22).

To determine the two bound state energies for the well.

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