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Q12P

Expert-verifiedFound in: Page 1215

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?**

- The longest wavelength is 68.6 nm .
- The third longest wavelength is 13.72 nm .

**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.**

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, $h=6.626\times {10}^{-34}\mathrm{Js}$

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

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

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

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

$f=\frac{\u2206E}{h}\phantom{\rule{0ex}{0ex}}=\left(\frac{{h}^{2}}{8m{L}^{2}}\right)\left(\frac{1}{h}\right)\left({n}_{f}^{2}-{n}_{i}^{2}\right)$

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.

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

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

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

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 $68.6\mathrm{nm}$ .

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 $25.72\mathrm{nm}$.

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 $13.72\mathrm{nm}$.

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