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Expert-verified Found in: Page 1215 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # An electron is trapped in a one-dimensional infinite well of width${\mathbf{250}}{\mathbf{pm}}$ and is in its ground state. What are the (a) longest, (b) second longest, and (c) third longest wavelengths of light that can excite the electron from the ground state via a, single photon absorption?

1. The longest wavelength is 68.6 nm .
2. The second longest wavelength is 25.72 nm .
3. The third longest wavelength is 13.72 nm .
See the step by step solution

## Step 1: Introduction

An electron is a negatively charged subatomic particle. It can be either free (not attached to any atom), or bound to the nucleus of an atom. Electrons in atoms exist in spherical shells of various radii, representing energy levels. The larger the spherical shell, the higher the energy contained in the electron.

## Step 2: Concept

An infinite potential well is a device for confining an electron. From the confinement principle we expect that the matter wave representing a trapped electron can exist only in a set of discrete states.

Energy of electron in an energy level in one-dimensional potential well is,

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

Here, h is the Planck’s constant, L is the length of the well, and m is the mass of the electron, and n is the integer.

Planck’s constant, $h=6.626×{10}^{-34}\mathrm{Js}$

Mass of the electron, $m=9.109×{10}^{-31}\mathrm{kg}$

Convert the length of the well from Pico meters to meters as follows:

$L=250\mathrm{pm}\phantom{\rule{0ex}{0ex}}=\left(250\mathrm{pm}\right)\left({10}^{-12}\mathrm{m}/\mathrm{pm}\right)\phantom{\rule{0ex}{0ex}}=250×{10}^{12}\mathrm{m}$

## Step 3: Calculation

When electron jumps from $n={n}_{1}$ state to $n={n}_{f}$ state by absorbing a

Photon, then the frequency of the photon is given by

$f=\frac{∆E}{h}\phantom{\rule{0ex}{0ex}}=\left(\frac{{h}^{2}}{8m{L}^{2}}\right)\left(\frac{1}{h}\right)\left({n}_{f}^{2}-{n}_{i}^{2}\right)$

Wavelength of the photon, $\lambda =\frac{c}{f}$

Substitute $\left(\frac{{h}^{2}}{8m{L}^{2}}\right)\left(\frac{1}{h}\right)\left({n}_{f}^{2}-{n}_{i}^{2}\right)$ for f in the above equation.

Here, the electron is the ground state is ${n}_{i}=1$

Hence, the wavelength is

$\lambda =\frac{8m{L}^{2}}{h\left({n}_{f}^{2}-{n}_{i}^{2}\right)}$

Substitute $9.10×{10}^{-31}\mathrm{kg}\mathrm{for}m,250×{10}^{-12}\mathrm{m}\mathrm{for}L,2.998×{10}^{8}m/s$ for c , $6.626×{10}^{-34}\mathrm{j}.\mathrm{s}$ for h , and 1 for ${n}_{i}$ in the above equation.

$\lambda =\frac{8\left(9.10×{10}^{-31}\mathrm{kg}\right){\left(250×{10}^{-12}\mathrm{m}\right)}^{2}\left(2.998×{10}^{8}\mathrm{m}/\mathrm{s}\right)}{\left(6.626×{10}^{-34}\mathrm{j}.\mathrm{s}\right)\left({n}_{f}^{2}-{1}^{2}\right)}\phantom{\rule{0ex}{0ex}}=\left(\frac{2.058×{10}^{7}\mathrm{m}}{{n}_{f}^{2}-1}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{2.058×{10}^{-9}\mathrm{m}\left({10}^{9}\mathrm{nm}/\mathrm{m}\right)}{{n}_{f}^{2}-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{205.8\mathrm{nm}}{{n}_{f}^{2}-1}$

## Step 4: (a) Determine the longest wavelength

For longest wavelength, ${n}_{f}=2$

The longest wavelength is,

$\lambda =\frac{205.8\mathrm{nm}}{{n}_{f}^{2}-1}$

Substitute 2 for in the above equation.

role="math" localid="1661859531615" $\lambda =\frac{205.8\mathrm{nm}}{{2}^{2}-1}\phantom{\rule{0ex}{0ex}}=68.6\mathrm{nm}$

Therefore, the longest wavelength is $68.6\mathrm{nm}$ .

## Step 5: (b) Determine the second longest wavelength.

For second longest wavelength, ${n}_{f}=3$

The second longest wavelength is,

$\lambda =\frac{205.8\mathrm{nm}}{{n}_{f}^{2}-1}$

Substitute 3 for ${n}_{f}$ in the above equation.

$\lambda =\frac{205.8\mathrm{nm}}{{3}^{2}-1}\phantom{\rule{0ex}{0ex}}=25.72\mathrm{nm}$

Therefore, the second longest wavelength is $25.72\mathrm{nm}$.

## Step 6: (c) Determine third longest wavelength

For third longest wavelength, ${n}_{f}=4$

The third longest wavelength is,

$\lambda =\frac{205.8\mathrm{nm}}{{n}_{f}^{2}-1}$

Substitute 4 for ${n}_{f}$ in the above equation.

$\lambda =\frac{205.8\mathrm{nm}}{{4}^{2}-1}\phantom{\rule{0ex}{0ex}}=13.72\mathrm{nm}$

Therefore, the third longest wavelength is $13.72\mathrm{nm}$. ### Want to see more solutions like these? 