26.Use Russell-Saunders LS coupling to derive the ground state levels of (a) Ne
(b) F (c) Sc (d) Zr (e) C.
30.
a. Construct a Grotrian diagram for the He emission spectrum (shown on
the right) and use it to assign as much of this spectrum as you can
using the data in the table below.
b. What is the energy difference between the 3
S (1s1
2s1
) and 1
S (1s1
2s1
)
levels? Why are these levels not the same energy, and why is the 3
S
(1s1
2s1
) level lower than the 1
S (1s1
2s1
) level, and not vice versa?
c. What is the energy difference between the 3
PJ (1s1
2p1
) and 1
P (1s1
2p1
)
levels? Compare your answer to your answer for part (c) and explain
any differences.
We can't draw diagrams/structures on this forum and most of the answers to your questions can't be done easily on this site. If you can rephrase or reconstruct them perhaps we can help.
What are the Russell-Saunders LS coupling derivations of the ground state levels of (a) Ne (b) F (c) Sc (d) Zr (e) C.
Configuration Level DE (cm-1)
1s2 1S 0.0
1s1 2s1 3S 159855.97
1S 166277.44
1s1 2p1 3P2 169086.76
3P1 169086.84
3P0 169087.83
1s1 2p1 1P 171134.89
1s1 3s1 3S 183236.79
1S 184864.82
1s1 3p1 3P2 185564.56
3P1 185564.58
3P0 185564.85
1s1 3d1 3D3 186101.54
3D2 186101.54
3D1 186101.59
1s1 3d1 1D 186104.96
1s1 3p1 1P 186209.36
1s1 4p1 1P 191492.71
To derive the ground state levels using Russell-Saunders (LS) coupling, you need to follow the Hund's rules:
1. Identify the electron configuration for each element.
2. Determine the total angular momentum quantum number (J) for each electron configuration.
3. Assign the ground state levels based on the values of J.
(a) Ne: The electron configuration for Ne is 1s² 2s² 2p⁶. Since the outermost electrons are in the p orbital, the total angular momentum quantum number (J) will be 1/2. Thus, the ground state level for Ne is 2P.
(b) F: The electron configuration for F is 1s² 2s² 2p⁵. Again, since the outermost electrons are in the p orbital, J will be 1/2. Therefore, the ground state level for F is 2P.
(c) Sc: The electron configuration for Sc is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹. In this case, we consider the 3d¹ electron as a localized electron with a total angular momentum quantum number of 2. The outermost 4s² electrons will also have J = 0. Thus, the ground state level for Sc is 2D.
(d) Zr: The electron configuration for Zr is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d². The three outermost electrons are in 5s² and 4d² orbitals. The 5s² electrons will have J = 0, and the 4d² electrons can have J = 2 or 4. Therefore, the ground state level for Zr is a combination of 2D and 4F.
(e) C: The electron configuration for C is 1s² 2s² 2p². The two outermost electrons are in the p orbital, so J will be 1. Thus, the ground state level for C is 3P.
For question 30, let's break it down:
(a) To construct the Grotrian diagram, you need to plot all the energy levels for the He emission spectrum and label each level according to its corresponding electron transition. The data in the table will provide you with the transitions and their corresponding wavelengths.
(b) The energy difference between the 3S (1s¹ 2s¹) and 1S (1s¹ 2s¹) levels can be calculated using the equation ΔE = hc/λ, where ΔE is the energy difference, h is Planck's constant, c is the speed of light, and λ is the wavelength. The levels are not the same energy because they correspond to different electron configurations. The 3S (1s¹ 2s¹) level is lower than the 1S (1s¹ 2s¹) level because the 3S level has a higher principal quantum number (n=3), indicating higher energy.
(c) To determine the energy difference between the 3PJ (1s¹ 2p¹) and 1P (1s¹ 2p¹) levels, you can use the same equation ΔE = hc/λ. Compare your answer to part (b) and explain any differences. The difference will arise from the fact that these transitions involve different electron configurations and, therefore, have different energy levels. The difference in energy levels can be attributed to the differences in the principal quantum numbers (n) and azimuthal quantum numbers (l) of the involved electron orbitals.