Question

Summarize the differences and similarities between different energy levels in a quantum oscillator. Specifically for the first two levels in figure 8.26, compare the angular frequency $\sqrt{{\mathbf{K}}_{\mathbf{s}}\mathbf{/}\mathbf{m}}$, the amplitude , and the kinetic energy$\mathit{k}$ at the same value of . ( In a quantum-mechanical analysis the concepts of angular frequency and amplitude require reinterpretation. Nevertheless, there remain elements of the classical picture. For example, larger amplitude corresponds to a higher probability of observing a large stretch.)

Short answer

The direction of the conventional current in the wire is in opposite direction.

The number of levels above the ground state of the spring oscillator is $2\times {10}^{31}$

Step-by-step solution

Determine the quantized vibrational energy levels for an atomic harmonic oscillator.

The quantized vibrational energy levels for an atomic harmonic oscillator are given by,

$E_N=Nh{\omega }_0+E_0$

Here,

$N$ is the principal quantum number.

$E_0$ is the ground state energy of harmonic oscillator.

${\omega }_0$ is the angular frequency.

$h$ is Planck’s constant.

$k_s$ is interatomic spring stiffness.

$m$ is mass of an atom.

So the energy of a spring-mass oscillator is given by,

$E=\frac{1}{2}{k}_s{A}^2$

where $A$ is amplitude of oscillation.

Determine the number of levels above the ground state of the spring mass oscillator.

On the equating the equation $E_N=Nh{\omega }_0+E_0$ and$E=\frac{1}{2}{k}_s{A}^2$ . We get,

$\begin{array}{rcl}Nh{\omega }_0+E_0& =& \frac{1}{2}{k}_s{A}^2\\ Nh{\omega }_0+\frac{1}{2}h{\omega }_0& =& \frac{1}{2}{k}_s{A}^2\\ \left(N+\frac{1}{2}\right)h{\omega }_0& =& \frac{1}{2}{k}_s{A}^2\\ \left(N+\frac{1}{2}\right)h\sqrt{\frac{k_s}{m}}& =& \frac{1}{2}{k}_s{A}^2\\ & & \end{array}$

By simplifying we get,

$\begin{array}{rcl}N+\frac{1}{2}& =& \frac{\frac{1}{2}{k}_s{A}^2}{\left(h\sqrt{\frac{k_s}{m}}\right)}\\ N& =& \frac{\frac{1}{2}{k}_s{A}^2}{\left(h\sqrt{\frac{k_s}{m}}\right)}-\frac{1}{2}\\ & & \end{array}$

On substituting the known values on the above equation. We get,

$\begin{array}{rcl}N& =& \frac{\frac{1}{2}\left(0.7\right){\left(0.2\right)}^2}{\left(\left(1·05\times {10}^{-34}\right)\sqrt{\frac{0.7}{0.02}}\right)}-\frac{1}{2}\\ & =& 2\times {10}^{31}\\ & & \end{array}$

The number of levels above the ground state of the spring oscillator is $2\times {10}^{31}$