Question

Consider a quantum harmonic oscillator.

a. What happens to the spacing between the nodes of the wave function as |x| increases? Why?

b. What happens to the heights of the antinodes of the wave function as |x| increases? Why?

c. Sketch a reasonably accurate graph of the n=8 wave function of a quantum harmonic oscillator.

Short answer

Electron's quantum state after partition is$n=6$

Step-by-step solution

Step 1. Given information

Energy of particle confined in a box of length $L$=

$E_n=\frac{n^2{h}^2}{8m{L}^2}$

Here,

$E_n$= energy of $n^{th}$state,

$h$= plank's constant,

$m=$mass of particle,

$L=$length of box and

$n$ denotes the state of particle (harmonics) (n=1,2,3 ...)

Step 2. Finding the electron's quantum state

From the law of conservation of energy,

the energy in $4nm$section should be equal to the energy in $12nm$section.

For $4nm$

$E_2=\frac{(2{)}^2{h}^2}{8m(4\mathrm{nm}{)}^2}$

For $12nm$

$E_n=\frac{n^2{h}^2}{8m(12\mathrm{nm}{)}^2}$

$\frac{(2{)}^2{h}^2}{8m(4\mathrm{nm}{)}^2}=\frac{n^2{h}^2}{8m(12\mathrm{nm}{)}^2}$

$\frac{(2{)}^2}{(4\mathrm{nm}{)}^2}=\frac{n^2}{(12\mathrm{nm}{)}^2}$

$n^2=\frac{\left(2{)}^2\right(12\mathrm{nm}{)}^2}{(4\mathrm{nm}{)}^2}$

$n=6$

Thus, the electron's quantum state after partition =$6$