Q1. Explain the effect of non-coulomb central potential on the electronic energy levels of an atom.

Q2. What is the ground state configuration of Ne.

Neon : 1s2^{2}s2^{2}p^{6}

Q3. What is the total degeneracy of the ground state of carbon.

Configuration : 1s^{2}2s^{2}2p^{2}

Degeneracy of a shell (g) = Y! / X!(Y - X)!

Where X is no. of electrons and Y is total no. of electron occupy in that shell.

For 1s X = 2, Y = 2 ; g = 1
For 2s X = 2, Y = 2 ; g = 1
For 2p X = 2, Y = 6 ; g = 15
So, G = 1*1*15 = 15.

Q4. Write one of the excited state configurations of nitrogen and calculate the degeneracy of the excited state.

Excited state of Nitrogen : 1s^{2}2s^{2}2p^{2}3p^{1}

So, total degeneracy is 90.

Q5. Write the ground state configuration of Fe (Z = 26); Ru (Z = 44) and Au (Z = 79). Discuss why it is different from K (Z = 19).

Sol. Configuration of Fe (Z = 26) = 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6} or [Ar]4s^{2}3d^{6}

Ru (Z = 44) = 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{6} or [Kr]5s^{2}4d^{6}

Au (Z = 79) = [Xe]^{4}f145d^{10}6s^{1}

K (Z = 19) = [Ar]4s^{1}

Q6. How many transition lines do you find for 2D → 2P transition of Ca+ after L.S. splitting. Show the energy level diagram and transition also.

Sol.

where A_{1} is the spin orbit compiling constant of 2D

where A_{2} is the spin orbit compiling constant of 2P

Now we can constant the enhancing level diagram with the transition selection scale

Q7. Same as Q.6 for ^{3}P → ^{3}D.

Q8. Write the coupled function | j_{1} j_{2} j m> = | 3 2 4 3> as a linear combination of uncoupled functions | j_{1} j_{2} m_{1} m_{2}> by calculating the appropriate Clebsch – Gorden coefficients.

Using

And we have to calculate |3243>

We can write maximum value of J

J=5

So,

Operating J- on both side, |54> can be determined.

Again using orthonormality and orthonormality relationship, |44> can be determined
Now appling J – on |44> one gets |43>,

Q9. Find out the terms arising from p_{2} and d_{2} configuration in j – j coupling scheme. Draw the relative energy level diagram and compare it with the relative energy level positions of the terms arising from L – S coupling.

For the equivalent electrons such as p2configuration

l_{1} = 1, s_{1} = ½ ; j_{1} = 3/2, 1/2.

l_{2} = 1, s_{2} = ½ ; j_{2} = 3/2, ½

J = 3,2,1,0.
{3/2,3/2}3, {3/2,3/2}2, {3/2,3/2}1, {3/2,3/2}0, à out of that for equivalent electrons only
and will exist.

.

Q10. Using M_{L} – M_{S} table, find out the terms arising from ppp configuration in LS coupling approximation.

Terms arising from three electron configuration ppp;

Terms arising from maximum M_{L} = 2 and maximum M_{S} = 1/2. Total combination for this term is 10. Take out from each block of M_{L} = 2,1,0,-1,-2 and M_{S} = 1⁄2 and -1⁄2 .
That reduce to

Next Terms arising from maximum M_{L} = 1 and maximum M_{S} = 1/2 is2P . Total combination for this term is 6. Take out from each block of ML = 1,0,-1, M_{S} = 1⁄2 and -1⁄2 .

Next Terms arising from maximum M_{L} = 0 and maximum M_{S} = 3/2 is^{4}S. Total combination for this term is 4. Take out from each block of ML = 0 M_{S} = 3/2, 1/2, -1/2, -3/2.

Q11. Two angular momenta J_{1}= 1 and J2= ½ are coupled to form a new angular momentum J. An operator V operates only on basis function | J_{1} m_{1}> not on | J_{2} m_{2}>. The values of the matrix elements of this operator in the uncoupled basis set | J_{1} m_{1}> are as follows.
< J_{1}= 1, m_{1} = 0| V | J_{1}= 1, m_{1} = 0 > = 3/2

Calculate the value of the matrix element < J= 1/2, M= 1/2| V | J= 1/2, M= 1/2 > .

J_{1} = 1, J_{2} = ½

J = J_{1} + J_{2} to |J_{1} - J2 |

So, J = 3/2, ½

For J = 3/2, m = 3/2, ½, -1/2, -3/2.

For J = ½, m = ½, -1/2.

We can write, | 3/2, 3/2 > = | 1,1 >| 1/2,1/2 >

Apply J- on | 3/2, 3/2 >,

taking ‘-‘ sign.

Q12. The fine structure of ^{4}D → ^{4}P transition for C^{+} is recorded. The transitions corresponding to ^{4}D_{3/2} → ^{4}P_{5/2} , ^{4}D_{5/2} → ^{4}P_{5/2} and ^{4}D_{1/2} → ^{4}P_{3/2} are observed at 6812.2 Å, 6800.5 Å and 6798.0 Å respectively.
(a) Calculate the spin – orbit constants ( in cm^{-1}) for ^{4}4D and ^{4}P states.
(b) Show all other allowed transitions on energy level diagram and calculate their wavelength (in Å).

Diagram,

A_{1} and A_{2} are spin – orbit constants of ^{4}D and ^{4}P respectively.

Calculate E_{0}, A_{1} and A_{2} from above three equations.

(b)

So energy level diagram is

Subtracting (1) from (2)

Also in (3)

Subtracting (4) from (5)

Q13. The normalized hydrogen atom ground state wave function is given by
Ψ_{100}(r) = (1/4π)^{1/2}(z/a_{0})^{3/2}2exp(-zr/a_{0}).
What is the ground state energy of Li^{+}, according to independent particle model. Considering the electron – electron repulsion, calculate the ground state energy of Li^{+} according to first order perturbation theory.

Li^{+} (z = 3), but it is like Helium atom having 2e^{-} in outermost orbit (1s^{2}).

From independent particle model, E = -2z^{2}E_{h}

E_{h} = -13.6eV

E = -2(3)^{2}(-13.6) eV = 244.8 eV

Q14. The transition ^{2}P_{1/2} → ^{2}P_{1/2} for an atom A is at 5500 Å. The same transition for another atom B is at 5501 Å. Calculate the minimum magnetic field for which the photons emitted by atom A will be absorbed by atom B.

Diagram,

Splitting in Zeeman pattern, ∆E = g_{J}μ_{B}Bm_{J}

For ^{2}P_{1/2}, g_{J} = 2/3 and m_{J} = ½ this gives, ∆E = (1/3) μ_{B}B

For ^{2}S_{1/2}, g_{J} = 2 and m_{J} = ½ this gives, ∆E = μ_{B}B