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Q11P

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 potential well. For what (a) higher quantum number and (b) lower quantum number is the corresponding energy difference equal to the energy of the n = 5 level? (c) Show that no pair of adjacent levels has an energy difference equal to the energy of the n = 6 level.**

- The higher the quantum number is 13 .
- The lower quantum number is 12 .
- This transition is not possible.

**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.**

The energy of the electron is related to the quantum number (n) , length of the potential well (L) is,

${E}_{n}=\left(\frac{{h}^{2}}{8m{L}^{2}}\right)$

Let the quantum number of higher order are n + 1 and n respectively.

Now from equation (1) energy of the n + 1 quantum number state is,

${E}_{n+1}=\left(\frac{{h}^{2}}{8mL}\right){\left(n+1\right)}^{2}\phantom{\rule{0ex}{0ex}}={E}_{1}{\left(n+1\right)}^{2}\left\{Sinceenergyofstaten=1is{E}_{n}=\left(\frac{{h}^{2}}{8m{L}^{2}}\right)\right\}$

From equation (1) energy of the quantum number state is,

${E}_{n}=\left(\frac{{h}^{2}}{8mL}\right){\left(n+1\right)}^{2}\phantom{\rule{0ex}{0ex}}={E}_{1}{\left(n\right)}^{2}\left\{Sinceenergyofstaten=1is{E}_{n}=\left(\frac{{h}^{2}}{8m{L}^{2}}\right)\right\}$

Then the energy corresponds to levels n = 5 is,

role="math" localid="1661773853907" ${\mathrm{E}}_{5}={\mathrm{E}}_{1}{\left(5\right)}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{5}={25}^{2}$ ….. (1)

Now the difference between energy levels is given by,

${E}_{n+1}-{E}_{n}=\left[(n+1{)}^{2}-{n}^{2}\right]{E}_{1}\phantom{\rule{0ex}{0ex}}=\left[\left({n}^{2}+1+2n\right)-{n}^{2}\right]{E}_{1}\phantom{\rule{0ex}{0ex}}{E}_{n+1}-{E}_{n}=\left(2n+1\right){E}_{1}$ ….. (2)

From the given data energy difference is equal to energy of the n = 5 level.

Then, by comparing equations (1) and (2), you have

$\begin{array}{l}\left(2n+1\right)E=25{E}_{1}\\ 2n+1=25\\ 2n=24\\ n=12\end{array}$

The higher energy level corresponding to quantum number is define by,

$n+1=12+1\phantom{\rule{0ex}{0ex}}=13$

Hence, the higher quantum number is 13.

The lowest energy levels corresponds to quantum number is define by,

n = 12

** **

Hence, the lower quantum number is 12.

The adjacent level has an energy difference is,

${E}_{n+1}-{E}_{n}={E}_{6}$

Substitute $(2n+1){\mathrm{E}}_{1}$ for $\left({E}_{n+1}-{E}_{n}\right)$ and ${E}_{1}{\left(6\right)}^{2}$ for ${E}_{6}$ n the above equation.

$\left(2n+1\right){E}_{1}={E}_{1}{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=36{E}_{1}$

$\begin{array}{l}2n+1=36\\ 2n=35\\ n=17.5\end{array}$

Hence, the value of quantum number must be an integer. So this transition no possible.

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