Atom Spectra and Term Symbols
a) What is the wavelength of an emission from the hydrogen atom due to transfer of an
electron from a dorbital for the principal quantum number n=4 to the lowest energy
p orbital? What wavelength region does the emission correspond to?
b) What possible term symbols are there for the H atom in the two states mentioned in
a)?
c) Calculate
i)
Wavelength limits (nm) for the absorption spectrum of Li2+(g) and
ii)
The ionization potential (kJ/mol) of the ion.
iii)
What wavelength regions does the absorption spectrum correspond to?
Solutions:
a)
We use the formula:
1
1
𝐸 = −𝑅𝐻 ∗ ( 2 − 2 )
𝑛𝑖 𝑛𝑓
Where ni = 4, nf = 2 and RH = 2.17 J.
1
1
𝐸 = −2.1677 ∗ 10−18 𝐽 ∗ ( 2 − 2 ) = 4.0644 ∗ 10−19 𝐽
4
2
𝑚
3.00 ∗ 108 𝑠
𝑐
𝜆 = ℎ ∗ = 6.626 ∗ 10−34 𝐽𝑠 ∗
= 𝟒𝟖𝟗. 𝟒 𝒏𝒎
𝐸
4.0644 ∗ 10−19 𝐽
In comparison, from NIST we acquire the wavenumber of the lowest P orbital and the
D orbital:
𝐸𝑝−𝑙𝑜𝑤 = 82.259 𝑐𝑚−1
𝐸𝑑−𝑙𝑜𝑤 = 102.824 𝑐𝑚−1
The difference in energy:
𝐸𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝐸𝑑−𝑙𝑜𝑤 − 𝐸𝑝−𝑙𝑜𝑤 = 102.824 𝑐𝑚−1 − 82.259 𝑐𝑚−1 = 20.565 𝑐𝑚−1
The wavelength of the emission is then:
𝜆=
1
𝐸𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 [𝑐𝑚−1 ]
= 4,8626 ∗ 10−5 𝑐𝑚 = 𝟒𝟖𝟔, 𝟐𝟔 𝒏𝒎
We see that there are some descrepancies in the outcomes of the wavelength but for
now they will do.
This emission belongs to the visible spectrum.
b) Electron in d-orbital:
We want to find the term symbols, 2s+1XJ, where s is the spin of the system and J is the
rotation.
Now, since only one electron resides in the d-orbital, s=1/2 :
1
2𝑠 + 1 = 2 ∗ + 1 = 𝟐
2
Furthermore, being in a d-orbital tells us that the angular momentum quantum
number, l, can be:
𝑙 = −2, −1, 0, +1, +2
Hund‘s rule now tells us that we should choose the highest possible value for l, i.e.
+2.
l
Term
Symbol - X
0
S
1
P
2
D
3
F
4
G
Etc.
Etc.
So from Hund‘s rule we find that our X is equal to D.
J is given by the equation:
𝐽 = |𝐿 + 𝑠|, … , |𝐿 − 𝑠|
𝐽 = 2+
1
1
𝟓 𝟑
,2 − = ,
2
2
𝟐 𝟐
So we can now round up the possible term symbols for the d-orbital.
2
𝐷5 , 2𝐷3
2
2
Using these exact steps, we find the term symbols for the p-orbital to be:
2
𝑃3 , 2𝑃1
2
2
c) i) Since Li2+ only has one electron we can expand Rydberg‘s formula, which was used
in a).
1
1
𝐸𝑙𝑜𝑤 = −𝑅𝐻 ∗ 𝑍 2 ∗ ( 2 − 2 ) , 𝑤ℎ𝑒𝑟𝑒 𝑍 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑛𝑢𝑐𝑙𝑒𝑢𝑠
𝑛𝑖 𝑛𝑓
1
1
𝐸𝑙𝑜𝑤 = −2,1677 ∗ 10−18 𝐽 ∗ 32 ∗ ( 2 − 2 ) = 1.4632 ∗ 10−17 𝐽
2
1
𝜆𝑙𝑜𝑤
𝑚
3.00 ∗ 108 𝑠
𝑐
−34
= ℎ ∗ = 6.626 ∗ 10 𝐽𝑠 ∗
= 13.6 𝑛𝑚
𝐸
1.4632 ∗ 10−17 𝐽
𝐸ℎ𝑖𝑔ℎ = −2,1677 ∗ 10−18 𝐽 ∗ 32 ∗ (
−34
𝜆ℎ𝑖𝑔ℎ = 6.626 ∗ 10
1
1
− 2 ) = 1.9509 ∗ 10−17 𝐽
2
∞
1
𝑚
3.00 ∗ 108 𝑠
𝐽𝑠 ∗
= 10.2 𝑛𝑚
1.9509 ∗ 10−17 𝐽
NIST however gives us:
𝐸𝑙𝑜𝑤 = 476035 𝑐𝑚−1 → 𝜆𝑙𝑜𝑤 = 21.0 𝑛𝑚
𝐸ℎ𝑖𝑔ℎ = 605691 𝑐𝑚−1 → 𝜆ℎ𝑖𝑔ℎ = 16.5 𝑛𝑚
So again we see that the Rydberg formula gives us a relatively crude assessment of
the energies of emission for hydrogen-like atoms.
ii) The ionization potential is simply the maximum absorbance possible for the Li2+
atom (given we use the Rydberg formula), when n ∞.
𝐼. 𝑃. = 1.9509 ∗ 10−17 𝐽 = 11,748
𝑘𝐽
𝑚ó𝑙
NIST however gives the ionization potential:
𝐼. 𝑃. = 610079 𝑐𝑚−1 = 7298.16
𝑘𝐽
𝑚ó𝑙
iii)The emission spectra for Li2+ is in the UV-spectrum but very close to the X-Ray limit (~10
nm).
Höfundur úrlausnar: Helgi Rafn Hróðmarsson, október, 2010