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Chapter 7: Q51E (page 282)

A wave function with a non-infinite wavelength-however approximate it might be- has nonzero momentum and thus nonzero kinetic energy. Even a single "bump" has kinetic energy. In either case, we can say that the function has kinetic energy because it has curvature- a second derivative. Indeed, the kinetic energy operator in any coordinate system involves a second derivative. The only function without kinetic energy would be a straight line. As a special case, this includes a constant, which may be thought of as a function with an infinite wavelength. By looking at the curvature in the appropriate dimension(s). answer the following: For a givenn,isthe kinetic energy solely

(a) radial in the state of lowest l- that is, l=0; and

(b) rotational in the state of highest l-that is, l=n-1?

Short Answer

Expert verified
  1. Yes, The kinetic energy is solely radial.
  2. No, Kinetic energy is not solely rotational.

Step by step solution

01

 Given data

Orbital angular momentum, l = 0.

Orbital angular momentum, l = n – 1.

02

 (a) To determine the kinetic energy solely radial in case of l = 0

The only angular solution to the l =0 state is a constant that has no curvature and thus has zero rotational kinetic energy. Hence, for any n, the state of lowest l must have only radial kinetic energy.

Thus, the kinetic energy is solely radial.

03

 To determine the kinetic energy solely rotational in case of l = n – 1

The radial solution, for a given n, to the l = n – 1 state is never constant, involves an exponential part, which has a curvature, and thus has nonzero radial kinetic energy.Hence, for any n, the state of highest l cannot be only rotational kinetic energy.

The wave function for the atom with the highest l still has radial dependence, so the kinetic energy is not solely rotational

No, kinetic energy is not solely rotational.

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Most popular questions from this chapter

Consider two particles that experience a mutual force but no external forces. The classical equation of motion for particle 1 is v˙1=F2on1/m1, and for particle 2 is v˙2=F1on2/m2, where the dot means a time derivative. Show that these are equivalent to v˙cm=constant, and v˙rel=FMutual/μ .Where, v˙cm=(m1v˙1+m2v˙2)/(m1m2),FMutual=-Fion2andμ=m1m2(m1+m2).

In other words, the motion can be analyzed into two pieces the center of mass motion, at constant velocity and the relative motion, but in terms of a one-particle equation where that particle experiences the mutual force and has the “reduced mass” μ.

To conserve momentum, an atom emitting a photon must recoil, meaning that not all of the energy made available in the downward jump goes to the photon. (a) Find a hydrogen atom's recoil energy when it emits a photon in a n = 2 to n = 1 transition. (Note: The calculation is easiest to carry out if it is assumed that the photon carries essentially all the transition energy, which thus determines its momentum. The result justifies the assumption.) (b) What fraction of the transition energy is the recoil energy?

Consider an electron in the ground state of a hydrogen atom. (a) Sketch plots of E and U(r) on the same axes (b) Show that, classically, an electron with this energy should not be able to get farther than 2a0from the proton. (c) What is the probability of the electron being found in the classically forbidden region?

Explicitly verify that the simple function Rr=Aebrcan be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.

Here we Pursue the more rigorous approach to the claim that the property quantized according to ml is Lz,

(a) Starting with a straightforward application of the chain rule,

∂∂φ = ∂x∂φ/∂∂x+∂y∂φ∂∂y+∂z∂φ∂∂z

Use the transformations given in Table 7.2 to show that

∂∂φ = -y∂∂x+x∂∂y

(b) Recall that L = r x p. From the z-component of this famous formula and the definition of operators for px and py, argue that the operator for Lz is -ih∂∂φ..

(c) What now allows us to say that our azimuthal solutioneimlφ has a well-defined z-component of angular momentum and that is value mlh.
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