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The harmonic oscillator potential energy is proportional to x2, and the energy levels are equally spaced:

En(n+12). The energy levels in the infinite well become farther apart as energy increases: Enn2.Because the functionlimb|x/L|bis 0 for|x|<Land infinitely large for|x|>L. the infinite well potential energy may be thought of as proportional to |x|.

How would you expect energy levels to be spaced in a potential well that is (a) proportional to |x|1and (b) proportional to -|x|-1? For the harmonic oscillator and infinite well. the number of bound-state energies is infinite, and arbitrarily large bound-state energies are possible. Are these characteristics shared (c) by the |x|1well and (d) by the-|x|-1well? V

Short Answer

Expert verified

(a) The energy levels are more closed for potential energy varies with -x-1.

(b) The energy levels are more closed for potential energy varies with-x-1.

(c) The number of level is infinite.

(d) The infinite number of level is possible.

Step by step solution

01

Step 1:Understanding the concept change in energy levels.

The separation between energy levels depends on the value of the polynomial function's exponent. In the case when the exponent is equal to 2, the separation is constant. However, when the exponent is higher than 2, the separation is relatively longer, and when the exponent is lower than 2, the energy separation is decreasing keeping them closer.

02

Part (a)Step 2: Applying the concept change in energy levels.

Draw a diagram to show the variation of potential energy.

The potential energy is varies as , in this case the energy level will be more closed as the energy increases as the potential energy decreases as the value of x increases.

Thus, the energy levels are more closed for potential energy varies with .

03

Part (b)Step 3: Applying the concept change in energy levels.

Draw a diagram to show the variation of potential energy.

The potential energy is varies as , in this case the energy level will be more closed as the energy increases as the potential energy decreases as the value of x increases.

Thus, the energy levels are more closed for potential energy varies with .

04

Understanding the concept of infinite potential well.

For infinite potential well the distance between the two levels is increased as the energy increases, as the energy is proportional to n2.

05

Part (c)Step 5: Applying the concept of infinite potential well.

The potential energy is varies as|x|-1, in this case the energy level will be more closed as the energy increases as the potential energy decreases as the value of x increases.

The potential well is infinitely deep, so the energy is high. As the number of level will be increased with the energy so the number of levels goes to infinity.

Thus, the number of level is infinite.

06

Part (d)Step 6: Applying the concept of infinite potential well.

The potential energy is varies as-|x|-1, in this case the energy level will be more closed as the energy increases as the potential energy decreases as the value of x increases.

The energy can not be much high, but the value of energy can not be greater than the potential energy. So the levels get vloser as the energy increases.

Thus, infinite number of level is possible.

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Most popular questions from this chapter

There are mathematical solutions to the Schr枚dinger 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 that(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 , 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-kA)andF=12e-L[(A-Bk)sin(kL)+(Ak+B)cos(kL)]

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

What is the product of uncertainties determined in Exercise 60 and 61? Explain.

Consider the delta well potential energy:

U(x)={0x0-x=0

Although not completely realistic, this potential energy is often a convenient approximation to a verystrong, verynarrow attractive potential energy well. It has only one allowed bound-state wave function, and because the top of the well is defined as U = 0, the corresponding bound-state energy is negative. Call its value -E0.

(a) Applying the usual arguments and required continuity conditions (need it be smooth?), show that the wave function is given by

(x)=(2mE0h2)1/4e-(2mE0/)|x|

(b) Sketch (x)and U(x) on the same diagram. Does this wave function exhibit the expected behavior in the classically forbidden region?

For a total energy of 0, the potential energy is given in Exercise 96. (a) Given these, to what region of the x-axis would a classical particle be restricted? Is the quantum-mechanical particle similarly restricted? (b) Write an expression for the probability that the (quantum-mechanical) particle would be found in the classically forbidden region, leaving it in the form of an integral. (The integral cannot be evaluated in closed form.)

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

sin(a+b)=sinacosb+cosasinband

cos(a+b)=cosacosb-sinasinbon the

functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that (z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as k=tankL2 and k=-cotkL2. Finally, plug these separately back into the region I solutions and show that

(z)=C{e(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z<L/2coskzcoskL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌-L/2<z<L/2e-(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z>L/2


or

(z)=C{e(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z<L/2-sinkzsinkL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌-L/2<z<L/2e-(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |(z)|2 over all space to 1 would give it in terms of kand , but because we can鈥檛 solve (22) exactly for k(or E), neither can we obtain an exact value for C.

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