Atoms are the smallest unit of matter. It acts as building blocks for everything. Atoms have various properties and are of great importance. Similarly, Properties of Atoms Spectra play a vital role in Physics. As the name implies, this chapter deals with the properties of atomic spectra. IE IRODOV Physics Solutions provides the solutions to various problems related to the concept. The questions are designed in such a manner that they cover the vital concepts and give detailed explanations to all the problems.
The solutions are based on the various formulae, equations and diagrams which come under this topic. It helps the students understand the concept and connects the theoretical concepts with practical problems. Students can easily understand and have a grip over the chapter. Students can easily face the upcoming board and competitive examination on solving and practising the problems. This helps in better understanding and building confidence in students.
1. The binding energy of a valence electron in a Li atom in the states 2S and 2P is equal to 5.39 and 3.54 eV respectively. Find the Rydberg corrections for S and P terms of the atom.
Solution:
From the Rydberg formula we know that
α_{1} = -0.41
For p state
α_{1} = -0.039
2. Find the Rydberg correction for the 3P term of a Na atom whose first excitation potential is 2.10 V and whose valence electron in the normal 3S state has the binding energy 5.14 eV.
Solution:
The energy of the 3p state must be -(E_{0}-e_{ϕ}) where - E_{0} is the energy of the 3S state.
Then
3. Find the binding energy of a valence electron in the ground state of a Li atom if the wavelength of the first line of the sharp series is known to be equal to λ_{1} = 813 nm and the short-wave cutoff wavelength of that series to λ_{2} = 350 nm.
Solution:
For the first line of the sharp series (3S→2P) in a Li atom
4. Determine the wavelengths of spectral lines appearing on transition of excited Li atoms from the state 3S down to the ground state 2S. The Rydberg corrections for the S and P terms are —0.41 and —0.04.
Solution:
We see that
E (2S ) <E (2P)< E ( 3 S)
The transitions are 3S → 2P and 2P →2S.
Direct 3S → 2S transition is forbidden by selection rules. The wavelengths are determined by
On substitution we get
λ = 0.816 μm ( 3S → 2P )
and λ = 0.674 μm ( 2P → 2S )
5. The wavelengths of the yellow doublet components of the resonance Na line caused by the transition 3P → 3S are equal to 589.00 and 589.56 nm. Find the splitting of the 3P term in eV units.
Solution:
The splitting of the Na lines is due to the fine structure splitting of 3 p lines (The 3 s state is nearly single except for possible hyperfine effects.) The splitting of the 3 p level then equals the energy difference
Here Δλ = wavelength difference & λ = average wavelength.
On substitution we get
ΔE = 2.0 meV
6. The first line of the sharp series of atomic cesium is a doublet with wavelengths 1358.8 and 1469.5 nm. Find the frequency intervals (in rad/s units) between the components of the sequent lines of that series.
Solution:
The sharp series arise from the transitions ns → mp . The s lines are unsplit so the splitting is due entirely to the p level. The frequency difference between sequential lines is
On evaluating we get
1.645 x 10^{14} rad/s
7. Write the spectral designations of the terms of the hydrogen atom whose electron is in the state with principal quantum number n = 3.
Solution:
We shall ignore hyperfine interaction. The state with principal quantum number n = 3 has orbital angular momentum quantum number / = 0, 1,2
The levels with these terms are 3 5, 3 P, 3 D. The total angular momentum is obtained by combining spin and angular momentum. For a single electron, this leads to
J=½ if L=0
J=L-½ and L+½ if L≠0
We then get the final designations like
^{3}S_{1/2}, ^{3}P_{1/2}, ^{3}P_{3/2}, ^{3}D_{3/2}, ^{3}D_{5/2}
8. How many and which values of the quantum number J can an atom possess in the state with quantum numbers S and L equal respectively to
(a) 2 and 3;
(b) 3 and 3;
(c) 5/2 and 2?
Solution:
The rule is that if
(a) The values are 1, 2, 3, 4, 5
(b) The values are 0, 1, 2, 3, 4, 5, 6
(c) The values are ½, 3/2, 5/2, 7/2, 9/2
9. Find the possible values of total angular momenta of atoms in the states 4P and 6D.
Solution:
For the state 4 p, L = 1, S = 3/2,(since 2s + 1 = 4 ) . For the state 5 d, Z, = 2 , s = 2 The possible values of J are
J:5/2, 3/2,1/2 for 4p
The value of the magnitude of angular momentum is h√J(J+1)
On substituting the values, we get
10. Find the greatest possible total angular momentum and the corresponding spectral designation of the term
(a) of a Na atom whose valence electron possesses the principal quantum number n = 4; (b) of an atom with electronic configuration 1s^{2}2p3d.
Solution:
(a) For the Na atoms the valence electron has principal quantum number n = 4, and the possible values of orbital angular momentum are l = 0 , 1, 2 , 3 so l_{max} = 3 . The state is ^{2}F, maximum value of J is 7/2
Thus the state with maximum angular momentum will be
For this state
(b) For the atom with electronic configuration 1 s^{2}2p 3d .There are two inequivalent valence electrons. The total orbital angular moments will be 1, 2, 3 so we pick l = 3. The total spin angular momentum will be s = 0, 1 so we pick up s = 1. Finally, 7 will be 2, 3, 4 so we pick up 4. Thus maximum angular momentum state for this state is
11. It is known that in F and D states the number of possible values of the quantum number J is the same and equal to five. Find the spin angular momentum in these states.
Solution:
For the f state L = 3, For the d state L = 2. Now if the state has spin s the possible angular momentum are |L-S| to L+S.
The number of J angular momentum values is 2 S + 1 if L ≥ S and 2 L + 1 if L < S.
Since the number of states is 5, we must have S ≥ L = 2 for D state while S ≤ 3 and 2 S + 1 = 5 in ply S = 2 for F state.
Thus for the F state total spin angular momentum
12. An atom is in the state whose multiplicity is three and the total angular momentum is
Solution:
Multiplicity is 2S + l so S = 1
Total angular momentum is
In order that J = 4 may be included in
|L-S| to L+S.
13. Find the possible multiplicities x of the terms of the types
(a) xD_{2}
(b) xHP_{3/2}
(c) xF_{1}
Solution:
(a) Here J = 2, L = 2. Then S = 0, 1 , 2, 3, 4 and the multiplicities( 2S - 1 ) are 1, 3, 5, 7, 9.
(b) Here J = 3 / 2 , L = 1 Then S= 5/2, 3/2, ½ and the multiplicities are 6, 4, 2
(c) Here J = 1, L = 3. Then S = 2 , 3 , 4 and the multiplicities are 5, 7, 9
14. A certain atom has three electrons (s, p, and d), in addition to filled shells, and is in a state with the greatest possible total mechanical moment for a given configuration. In the corresponding vector model of the atom find the angle between the spin momentum and the total angular momentum of the given atom.
Solution:
The total angular momentum is greatest when L, S are both greatest and add to form J. Now for a triplet of s, p> d electrons Maximum spin→S =3/2 corresponding to
15. An atom possessing the total angular momentum
Solution:
Total angular momentum
Thus E = 2 and the spectral symbol of the state is ^{3}D_{2}
16. Write the spectral symbols for the terms of a two-electron system consisting of one p electron and one d electron.
Solution:
In a system containing a p electron and a d electron S = 0,1 L = 1,2,3
For S = 0 we have the terms
^{1}P_{1}, ^{1}D_{2}, ^{1}F_{3}^{}
For S = 1 we have the terms
^{3}P_{0} , ^{3}P_{1}, ^{3}P_{2}, ^{3}D_{1}, ^{3}D_{2}, ^{3}D_{3}, ^{3}F_{1}, ^{3}F_{2}, ^{3}F_{3}
17. A system comprises an atom in ^{2}P_{3/2 } state and a d electron. Find the possible spectral terms of that system.
Solution:
$_{1}= ½, l_{1}=1, j = 3/2
The electron has $_{2}= ½, l_{2}=2 so the total angular momentum quantum number must j_{2}=3/2 or 5/2.
In L - S compiling we get S = 0, 1 . L = 1,2, 3 and the terms that can be formed are the same as written in the problem above. The possible values of angular momentum are consistent with the addition j_{1}= 3/2 0r j_{2}= 3/2 or 5/2.
The latter gives us J= o, 1, 2, 3; 1,2,3,4
All these values are reached above
18. Find out which of the following transitions are forbidden by the selection rules:
^{2}D_{3/2}→^{2}P_{1/2}, ^{3}P_{1}→^{2}S_{1/2}, ^{3}F_{3}→^{3}P_{2}, ^{4}F_{7/2}→^{4}D_{5/2}
Solution:
Selection rules are ΔS = 0
Thus,
^{2}D_{3/2}→^{2}P_{1/2 } is allowed
^{3}P_{1}→^{2}S_{1/2 }is allowed
^{3}F_{3}→^{3}P_{2 }is not allowed
^{4}F_{7/2}→^{4}D_{5/2 }is allowed
19. Determine the overall degeneracy of a 3D state of a Li atom. What is the physical meaning of that value?
Solution:
For a 3 d state of a Li atom, S= ½ because there is only one electron and L = 2
The total degeneracy is g= (2L + 1) (2S + 1) = 5 x 2 = 10.
The states are ^{2}D_{3/2} and ^{2}D_{5/2}. and we check that
g = 4 + 6 = {2 X 3/2+1} {2 X 3/2+1}
20. Find the degeneracy of the states ^{2}P, ^{3}D, and ^{4}F possessing the greatest possible values of the total angular momentum.
Solution:
The state with greatest possible total angular momentum are
For
^{2}P state
J= ½ + 1 = 3/2 i.e ^{2}P_{3/2}
Its degeneracy is 4.
For
^{3}D state
J= 1 + 2 = 3 i.e ^{3}D_{3}
Its degeneracy is 2 x 3 + 1=7
For
^{4}F state
J= 3/2 + 3 = p/2 i.e ^{4}F_{4/2}
Its degeneracy is 2 x 9/2 + 1=10.