An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (^mel = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 210 nm B) 200 nm C) 220 nm D) 230 nm

Question

Quizplus · Accepted Answer

The longest wavelength of light that the electron can absorb corresponds to the energy difference between two consecutive levels of the harmonic oscillator potential. The energy of a quantum harmonic oscillator is given by:

$E_n = \left(n + \frac{1}{2}ight)h
u$

where $n$ is the quantum number, $h$ is Planck's constant, and $
u$ is the frequency of the oscillator. The frequency of the oscillator is related to its spring constant $k$ by:

$
u = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where $m$ is the mass of the oscillator. In our case, the oscillator is an electron confined in a potential well, so $m$ is the electron mass. We are given that the net force on the electron behaves as though it has a spring constant of 74 N/m, so we can use this value to calculate the frequency of the oscillator:

$
u = \frac{1}{2\pi}\sqrt{\frac{74}{9.11	imes10^{-31}}} = 5.68	imes10^{15} \; 	ext{Hz}$

The energy difference between two consecutive levels of the oscillator is given by:

$\Delta E = E_{n+1} - E_n = h
u$

We want to find the longest wavelength of light that the electron can absorb, so we need to find the smallest energy difference. The smallest energy difference corresponds to the transition between the ground state ($n=0$) and the first excited state ($n=1$):

$\Delta E = E_1 - E_0 = \left(1 + \frac{1}{2}ight)h
u - \frac{1}{2}h
u = \frac{1}{2}h
u$

Substituting the value of $
u$ we found earlier, we get:

$\Delta E = \frac{1}{2}h
u = 4.24	imes10^{-20} \; 	ext{J}$

We can convert this energy to a wavelength using the formula:

$E = hc/\lambda$

where $\lambda$ is the wavelength of the light. Rearranging, we get:

$\lambda = hc/E$

Substituting the value of $\Delta E$ we found earlier, we get:

$\lambda = \frac{hc}{\Delta E} = \frac{hc}{4.24	imes10^{-20}} = 210 \; 	ext{nm}$

Therefore, the longest wavelength of light that the electron can absorb is 210 nm. The answer is A).

An Electron Is Confined in a Harmonic Oscillator Potential Well