An electron is trapped in an infinite square well (a box) of width 6.88 nm. Find the wavelength of photons emitted when the electron drops from the n = 5 state to the n = 1 state in this system. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34J ∙ s, ^mel = 9.11 × 10^-31 kg) A) 6.49 μm B) 5.45 μm C) 5.91 μm D) 7.07 μm

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Quizplus · Accepted Answer

The energy of an electron in an infinite square well is given by:$$E_n = \frac{n^2h^2}{8mL^2}$$where:* $n$ is the principal quantum number* $h$ is Planck's constant* $m$ is the mass of the electron* $L$ is the width of the wellThe wavelength of a photon emitted when an electron drops from a higher energy state to a lower energy state is given by:$$\lambda = \frac{hc}{E_i - E_f}$$where:* $h$ is Planck's constant* $c$ is the speed of light* $E_i$ is the initial energy of the electron* $E_f$ is the final energy of the electronIn this problem, the electron drops from the $n = 5$ state to the $n = 1$ state. The energy of the electron in the $n = 5$ state is:$$E_5 = \frac{5^2h^2}{8mL^2}$$The energy of the electron in the $n = 1$ state is:$$E_1 = \frac{1^2h^2}{8mL^2}$$The wavelength of the photon emitted is:$$\lambda = \frac{hc}{E_5 - E_1} = \frac{hc}{4\frac{h^2}{8mL^2}} = \frac{2mL^2c}{h}$$Substituting the given values into the equation, we get:$$\lambda = \frac{2(9.11 	imes 10^{-31} 	ext{ kg})(6.88 	imes 10^{-9} 	ext{ m})^2(3.00 	imes 10^8 	ext{ m/s})}{6.626 	imes 10^{-34} 	ext{ J s}} = 6.49 	imes 10^{-6} 	ext{ m} = 6.49 \mu	ext{m}$$Therefore, the wavelength of the photon emitted is 6.49 μm.

An Electron Is Trapped in an Infinite Square Well (A