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Chapter 5: Q 12CQ (page 186)

Simple models are very useful. Consider the twin finite wells shown in the figure, at First with a tiny separation. Then with increasingly distant separations, In all case, the four lowest allowed wave functions are planned on axes proportional to their energies. We see that they pass through the classically forbidden region between the wells, and we also see a trend. When the wells are very close, the four functions and energies are what we might expect of a single finite well, but as they move apart, pairs of functions converge to intermediate energies.

(a) The energies of the second and fourth states decrease. Based on changing wavelength alone, argue that is reasonable.

(b) The energies of the first and third states increase. Why? (Hint: Study bow the behaviour required in the classically forbidden region affects these two relative to the others.)

(c) The distant wells case might represent two distant atoms. If each atom had one electron, what advantage is there in bringing the atoms closer to form a molecule? (Note: Two electrons can have the same wave function.)

Short Answer

Expert verified

(a)The energies of the second and the fourth states decrease.

(b)The energies of the first and third states are larger.

(c) The advantage in bringing the atoms closer is to lower the energy of the state to form an atom.

Step by step solution

01

The reason of the decreasing nature of the second and the fourth states.

The separation of a twin well, initially having a tiny separation, increases. The four wave functions pass through the classical forbidden region between the walls. When the wells are very close, the four wave functions and energies are very similar to a single finite well, but as they move apart, pairs of functions converge to immediate energies.

The second and fourth have nodes in the middle, and simply assume a longer wavelength as the wells separate. A larger wavelength implies a smaller momentum and thus kinetic energy.

02

The reason of the increasing nature of the first and the third states.

The separation of a twin well, initially having a tiny separation, increases. The four wave functions pass through the classical forbidden region between the walls. When the wells are very close, the four wave functions and energies are very similar to a single finite well, but as they move apart, pairs of functions converge to immediate energies.

The first and third state energies are required to die off in the classically forbidden region, where they would naturally have an antinode. Thus, they assume effectively shorter wavelengths than in one single well. A shorter wavelength implies a larger value of kinetic energy.

03

The reason of the advantage in bringing the atoms closer to from a molecule.

The separation of a twin well, initially having a tiny separation, increases. The four wave functions pass through the classical forbidden region between the walls. When the wells are very close, the four wave functions and energies are very similar to a single finite well, but as they move apart, pairs of functions converge to immediate energies.

When an antinode is approached from middle it will generate a value of wavelength. So the energy corresponding to the antinode point is low to form a ground state.

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Most popular questions from this chapter

For the harmonic oscillator potential energy, U=12kx2, the ground-state wave function is 蠄(x)=Ae-(mk/2鈩�)x2, and its energy is 12鈩�k/m.

(a) Find the classical turning points for a particle with this energy.

(b) The Schr枚dinger equation says that 蠄(x) and its second derivative should be of the opposite sign when E > Uand of the same sign when E < U . These two regions are divided by the classical turning points. Verify the relationship between 蠄(x)and its second derivative for the ground-state oscillator wave function.

(Hint:Look for the inflection points.)

Obtain expression (5-23) from equation (5-22). Using 肠辞蝉胃=cos2(12胃)鈭�sin2(12胃)and蝉颈苍胃=2sin(12胃)cos(12胃), first convert the argument of the cotangent fromkLto12kL. Next, put the resulting equation in quadratic form, and then factor. Note that伪is positive by definition.

a) Taking the particle鈥檚 total energy to be 0, find the potential energy.

(b) On the same axes, sketch the wave function and the potential energy.

(c) To what region would the particle be restricted classically?

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蠄(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 伪, 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-k伪A)andF=12e-伪L[(A-Bk伪)sin(kL)+(Ak伪+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.
See all solutions

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