Question

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

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

Step-by-step solution

 Given data

Orbital angular momentum, l = 0.

Orbital angular momentum, l = n – 1.

 (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.

 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.