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? (m_el = 9.11 × 10^-31 kg,c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J,h = 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

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Quizplus · Accepted Answer

The longest wavelength of light that the electron can absorb is given by the equation:$$\lambda = \frac{hc}{E_n - E_0}$$where:* $\lambda$ is the wavelength of light* $h$ is Planck's constant* $c$ is the speed of light* $E_n$ is the energy of the electron in the excited state* $E_0$ is the energy of the electron in the ground stateThe energy of the electron in the ground state is given by:$$E_0 = \frac{1}{2}hv_0$$where:* $v_0$ is the frequency of the oscillatorThe energy of the electron in the excited state is given by:$$E_n = \frac{n+1}{2}hv_0$$where:* $n$ is the principal quantum numberThe net force on the electron is given by:$$F = -kx$$where:* $k$ is the spring constant* $x$ is the displacement of the electron from its equilibrium positionThe spring constant is given by:$$k = m\omega^2$$where:* $m$ is the mass of the electron* $\omega$ is the angular frequency of the oscillatorThe angular frequency is given by:$$\omega = \sqrt{\frac{k}{m}}$$Substituting the given values into the equations, we get:$$\lambda = \frac{hc}{\frac{1}{2}hv_0 - \frac{n+1}{2}hv_0}$$$$\lambda = \frac{2hc}{hv_0(n+1)}$$$$\lambda = \frac{2hc}{k(n+1)}$$$$\lambda = \frac{2(6.626 	imes 10^{-34} 	ext{ J s})(3.00 	imes 10^8 	ext{ m/s})}{74 	ext{ N/m}(1+1)}$$$$\lambda = 210 	ext{ nm}$$Therefore, the longest wavelength of light that the electron can absorb is 210 nm.

An Electron Is Confined in a Harmonic Oscillator Potential Well