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Q24P

Expert-verifiedFound in: Page 1248

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**For Problem 20, what multiple of ${{\mathbf{h}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{8}}{{\mathbf{mL}}}^{{\mathbf{2}}}$ gives the energy of (a) the first excited state, (b) the second excited state, and (c) the third excited state of the system of seven electrons? (d) Construct an energy-level diagram for the lowest four energy levels.**

a) The multiple value of ${\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}$ that gives the energy of the first excited state of the system is 18 .

b) The multiple value of ${\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}$ that gives the energy of the second excited state of the system is 18.25.

c) The multiple value of ${\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}$ that gives the energy of the third excited state of the system is 19.

d) The energy-level diagram for the lowest four energy levels is constructed.

A rectangular coral of widths, ${\mathrm{L}}_{\mathrm{x}}=\mathrm{L}\mathrm{and}{\mathrm{L}}_{\mathrm{y}}=2\mathrm{L}$

**The low state of the quantum-mechanical system is its fixed state of extremely low power; the power of the lower state is known as the zero power of the system. **

**Pauli's exclusive principle states that no two electrons in the same atom can have the same values in all four of their quantum numbers.**

Using Eq. 39-20, the lowest four levels of the rectangular coral are non-degenerate. Using the concept of Pauli's exclusion principle, the total energy of the non-degenerate levels in the multiple of our required value ${\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}$. Using similar values and the condition of the excited state, the electrons occupying each state; thus, the total energy in each excited state.

Using Eq. 39-20, the lowest five levels of the rectangular coral having energies is,

${\mathrm{E}}_{1,1}=1.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{1,2}=2.00{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{1,3}=3.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{2,1}=4.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{2,2}=5{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}$

Here, the energy level ${\mathrm{E}}_{2,2}$ corresponds to the two energy levels ${\mathrm{E}}_{2,2}\mathrm{and}{\mathrm{E}}_{1,4}$ that degenerate.

Now, for the case of the energy in the first excited state that is next lower to that of the ground state of the system, the total energy is as follow. (The configuration to this state consists of two electrons in the first three low energy levels, while the fourth is empty and the fifth being half-filled that is filled with one electron).

role="math" localid="1661920201557" ${\mathrm{E}}_{\mathrm{first}\mathrm{excited}}=2\left(1.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+2\left(2.00{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+2\left(3.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+\left(5.00{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)\phantom{\rule{0ex}{0ex}}=18\left({\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)$

Hence, the value of the multiple is 18 .

Now, for the case of the energy in the second excited state that is the next higher energy of the system, we can get the total energy as follows: (The configuration to this state consists of two electrons filled in the first two low energy levels, while the third one half-filled, and the fourth being full-filled by two electrons).

${\mathrm{E}}_{\mathrm{second}\mathrm{excited}}=2\left(1.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+2\left(2.00{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+\left(3.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+2\left(4.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)\phantom{\rule{0ex}{0ex}}=18.25\left({\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)$

Hence, the value of the multiple is 18.25 .

Now, for the case of the energy in the third excited state that is the next higher energy of the system, the total energy is as follow. (The configuration to this state consists of two electrons filled in the first two low energy levels, and next three levels half -filled).

${\mathrm{E}}_{\mathrm{third}\mathrm{excited}}=2\left(1.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+2\left(2.00{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+\left(3.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+2\left(4.25{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)+\left(5{\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)\phantom{\rule{0ex}{0ex}}=19\left({\mathrm{h}}^{2}/8{\mathrm{mL}}^{2}\right)$

Hence, the value of the multiple is 19 .

The energy states of this problem are given below in the diagram as:

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