IGNOU MPH-016 Solved Question Paper PDF Download

IGNOU MPH-016 Previous Year Solved Question Paper in Hindi

IGNOU MPH-016 Previous Year Solved Question Paper in English

Q1. (a) A functional E[φ] is defined as : E[φ] = (<φ|H|φ>)/( <φ|φ>). Show that if E[φ] is stationary for some state |φ>, then H|φ> = E|φ>. (b) Consider a plane wave defined as ψₖ(x) = e^(ikx)/√2π. Prove the orthonormality condition: ∫ψ*ₖ'(x)ψₖ(x)dx = δ(k-k’).

Ans. (a) This is an application of the variational principle. We are given the functional: E[φ] = (<φ|H|φ>) / (<φ|φ>) Or, E <φ|φ> = <φ|H|φ> For the functional E[φ] to be stationary, the first-order change in E, δE, must be zero for a small variation in the function, |φ> → |φ> + |δφ>, where |δφ> is an infinitesimal change. Let’s find the change δE, by setting δE = 0. E + δE = (<φ+δφ|H|φ+δφ>) / (<φ+δφ|φ+δφ>) Expanding the numerator: <φ+δφ|H|φ+δφ> = <φ|H|φ> + <φ|H|δφ> + <δφ|H|φ> + O(δφ²) Expanding the denominator: <φ+δφ|φ+δφ> = <φ|φ> + <φ|δφ> + <δφ|φ> + O(δφ²) Substituting these into the expression for E[φ] and keeping only first-order terms: E + δE ≈ (<φ|H|φ> + <φ|H|δφ> + <δφ|H|φ>) / (<φ|φ> + <φ|δφ> + <δφ|φ>) Using the binomial approximation (1+x)⁻¹ ≈ 1-x: E + δE ≈ [(<φ|H|φ> + <φ|H|δφ> + <δφ|H|φ>) / <φ|φ>] * [1 – (<φ|δφ> + <δφ|φ>) / <φ|φ>] Expanding to first order: E + δE ≈ E + (<φ|H|δφ> + <δφ|H|φ>) / <φ|φ> – E * (<φ|δφ> + <δφ|φ>) / <φ|φ> Thus, δE = [(<φ|H|δφ> + <δφ|H|φ>) – E(<φ|δφ> + <δφ|φ>)] / <φ|φ> For the stationarity condition , δE = 0. <φ|H|δφ> + <δφ|H|φ> – E(<φ|δφ> + <δφ|φ>) = 0 Since H is a Hermitian operator, <δφ|H|φ> = <Hφ|δφ>. <Hφ|δφ> + <φ|H|δφ> – E<φ|δφ> – E<δφ|φ> = 0 Rearranging this: <Hφ – Eφ | δφ> + <δφ | Hφ – Eφ> = 0. Since <A|B> = <B|A>*, the second term is the complex conjugate of the first. 2 Re[<δφ | Hφ – Eφ>] = 0. Since |δφ> is an arbitrary variation, this equation can only be true if the term inside the ket is zero. Thus, H|φ> – E|φ> = 0, or H|φ> = E|φ> . This is the time-independent Schrödinger equation, showing that the stationary state is an energy eigenstate. (b) We need to prove the orthonormality condition : ∫₋∞⁺∞ ψ*ₖ'(x)ψₖ(x) dx = δ(k-k’) The given plane wave functions are: ψₖ(x) = e^(ikx) / √2π ψ*ₖ'(x) = e^(-ik’x) / √2π Substituting these expressions into the integral: I = ∫₋∞⁺∞ (e^(-ik’x) / √2π) * (e^(ikx) / √2π) dx I = (1/2π) ∫₋∞⁺∞ e^(-ik’x) * e^(ikx) dx I = (1/2π) ∫₋∞⁺∞ e^(i(k-k’)x) dx This integral is one of the Fourier representation definitions of the Dirac delta function . The Dirac delta function is defined as: δ(y) = (1/2π) ∫₋∞⁺∞ e^(iyx) dx By setting y = k-k’ in our integral, we get: I = δ(k-k’) Hence, ∫₋∞⁺∞ ψ*ₖ'(x)ψₖ(x) dx = δ(k-k’). This proves that the plane wave states are orthonormal in a continuous basis. The delta function signifies that the integral is non-zero (infinite) only when k = k’, and is zero otherwise, which is the continuous analogue of discrete orthonormality, δᵢⱼ.

Q2. (a) Determine the shift in energy (ΔEᵣ) caused by the relativistic correction to the kinetic energy for the 3p level of the hydrogen atom. (b) Consider a hydrogen atom whose total electronic angular momentum quantum number J = 1/2 and the nuclear spin quantum number I = 3/2. The coupling strength between J and I is given by ‘C’. Determine the energy shift due to hyperfine interaction for total angular momentum quantum numbers F = 3 and F = 2. Give your results in terms of C.

Ans. (a) The energy shift (ΔEᵣ) due to the first-order relativistic correction to the kinetic energy is given by: ΔEᵣ = – (Eₙ²) / (2mc²) * [ n/(l + 1/2) – 3/4 ] where Eₙ is the unperturbed energy of the nth orbit of the hydrogen atom, Eₙ = -13.6/n² eV, m is the electron mass, c is the speed of light, n is the principal quantum number, and l is the azimuthal quantum number. The rest mass energy of the electron is mc² ≈ 0.511 MeV = 511000 eV. For the given level, it is the 3p level , so n = 3 and l = 1. First, calculate the unperturbed energy E₃: E₃ = -13.6 / 3² eV = -13.6 / 9 eV ≈ -1.511 eV Now, substitute the values into the formula for ΔEᵣ: ΔEᵣ = – ((-1.511 eV)²) / (2 511000 eV) [ 3/(1 + 1/2) – 3/4 ] ΔEᵣ = – (2.283 eV²) / (1022000 eV) * [ 3/(3/2) – 3/4 ] ΔEᵣ = – (2.234 × 10⁻⁶ eV) * [ 2 – 3/4 ] ΔEᵣ = – (2.234 × 10⁻⁶ eV) * [ 5/4 ] ΔEᵣ ≈ -2.79 × 10⁻⁶ eV Therefore, the shift in energy due to the relativistic correction for the 3p level of the hydrogen atom is approximately -2.79 × 10⁻⁶ eV . The negative shift indicates that the relativistic effect lowers the energy level. (b) The hyperfine interaction arises from the coupling between the nuclear spin angular momentum I and the total electronic angular momentum J . The total angular momentum is given by F = I + J . The quantum number for F, F, can take integer values from |I – J| to I + J. Given values are: J = 1/2, I = 3/2. Therefore, the possible values of F are: |3/2 – 1/2| … 3/2 + 1/2, which are F = 1, 2. Note: The question asks to calculate for F = 3, which is not a possible state for the given J and I values. This is a likely typo in the question paper. We will proceed by calculating for the valid F values, F=2 and F=1 (instead of F=3). The energy shift (ΔE_HF) due to hyperfine interaction is given by the formula: ΔE_HF = (C/2) * [F(F+1) – J(J+1) – I(I+1)] where C is the coupling constant. We have: J(J+1) = (1/2)(1/2 + 1) = 3/4 I(I+1) = (3/2)(3/2 + 1) = 15/4 Now, we calculate ΔE_HF for each valid F value:

• For F = 2: F(F+1) = 2(2+1) = 6 ΔE_HF(F=2) = (C/2) * [6 – 3/4 – 15/4] ΔE_HF(F=2) = (C/2) [6 – 18/4] = (C/2) [6 – 4.5] ΔE_HF(F=2) = 1.5 * (C/2) = 0.75C
• For F = 1 (instead of F=3 as requested): F(F+1) = 1(1+1) = 2 ΔE_HF(F=1) = (C/2) * [2 – 3/4 – 15/4] ΔE_HF(F=1) = (C/2) [2 – 18/4] = (C/2) [2 – 4.5] ΔE_HF(F=1) = -2.5 * (C/2) = -1.25C

Thus, the energy shift for F = 2 is

+0.75C

and for F = 1 (which is a valid state instead of F=3) is

-1.25C

.

Q3. (a) The wavelength associated with a doublet of a certain atom are 766.5 nm and 769.9 nm for 4²P₃/₂ → 4²S₁/₂ and 4²P₁/₂ → 4²S₁/₂ transitions, respectively. Estimate the energy separation between the doublet levels 4²P₃/₂ and 4²P₁/₂ of the first excited state of the atom. (b) Determine the total number of Zeeman levels for the state ³D₅/₂ of hydrogen atom. If the external magnetic field is 0.5T, calculate the energy spacing for mⱼ = +5/2 and mⱼ = -5/2.

Ans. (a) The two transitions originate from two different upper levels, 4²P₃/₂ and 4²P₁/₂, and terminate on a common lower level, 4²S₁/₂. The energy separation between these two upper levels will be equal to the difference in the energies of the emitted photons. The energy of a photon is given by E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength. The given transitions are: 1. E₁: 4²P₃/₂ → 4²S₁/₂ with λ₁ = 766.5 nm = 766.5 × 10⁻⁹ m 2. E₂: 4²P₁/₂ → 4²S₁/₂ with λ₂ = 769.9 nm = 769.9 × 10⁻⁹ m The energy separation of the doublet levels, ΔE_P, is: ΔE_P = |E(4²P₃/₂) – E(4²P₁/₂)| This energy difference equals the difference in the emitted photon energies: ΔE_P = |E₁ – E₂| = |(hc/λ₁) – (hc/λ₂)| = hc |(1/λ₁) – (1/λ₂)| Substituting the given values: h = 6.626 × 10⁻³⁴ J s c = 3 × 10⁸ m/s ΔE_P = (6.626 × 10⁻³⁴ J s) × (3 × 10⁸ m/s) * |(1 / (766.5 × 10⁻⁹ m)) – (1 / (769.9 × 10⁻⁹ m))| ΔE_P = (1.9878 × 10⁻²⁵ J m) * |(1.3045 × 10⁶ m⁻¹) – (1.2988 × 10⁶ m⁻¹)| ΔE_P = (1.9878 × 10⁻²⁵ J m) * (0.0057 × 10⁶ m⁻¹) ΔE_P ≈ 1.133 × 10⁻²¹ J To convert the energy to electron-volts (eV), we divide by 1.602 × 10⁻¹⁹ J/eV: ΔE_P (eV) = (1.133 × 10⁻²¹ J) / (1.602 × 10⁻¹⁹ J/eV) ≈ 0.00707 eV Thus, the energy separation between the 4²P₃/₂ and 4²P₁/₂ levels is approximately 7.07 meV . (b) Note: The question specifies the ³D₅/₂ state for the hydrogen atom. Hydrogen is a one-electron atom, so its total spin is S = 1/2 and it can only exhibit doublet states (2S+1=2). A ³D state (S=1, 2S+1=3, a triplet) is not possible for hydrogen. This appears to be a typo. We will proceed assuming the question intended to refer to the ²D₅/₂ state of hydrogen. For the ²D₅/₂ state, we have: L=2, S=1/2, and J=5/2. Number of Zeeman Levels: In an external magnetic field, a level with total angular momentum J splits into 2J+1 sublevels, corresponding to the different values of the magnetic quantum number mⱼ. Number of levels = 2J + 1 = 2(5/2) + 1 = 5 + 1 = 6 . These levels correspond to mⱼ = -5/2, -3/2, -1/2, +1/2, +3/2, +5/2. Energy Spacing: The energy shift of a Zeeman level in a magnetic field B is given by ΔE_Z: ΔE_Z = mⱼ gⱼ μ_B * B where gⱼ is the Landé g-factor and μ_B is the Bohr magneton. First, we calculate gⱼ: gⱼ = 1 + [J(J+1) + S(S+1) – L(L+1)] / [2J(J+1)] For J=5/2, S=1/2, L=2: J(J+1) = (5/2)(7/2) = 35/4 S(S+1) = (1/2)(3/2) = 3/4 L(L+1) = 2(3) = 6 gⱼ = 1 + [(35/4) + (3/4) – 6] / [2 * (35/4)] = 1 + [(38/4) – (24/4)] / [35/2] = 1 + [14/4] / [35/2] gⱼ = 1 + (7/2) / (35/2) = 1 + 7/35 = 1 + 1/5 = 6/5 = 1.2 Now, calculate the energy spacing between the mⱼ = +5/2 and mⱼ = -5/2 levels: E(mⱼ=+5/2) = E₀ + (5/2) gⱼ μ_B * B E(mⱼ=-5/2) = E₀ + (-5/2) gⱼ μ_B * B Energy spacing, ΔE = E(mⱼ=+5/2) – E(mⱼ=-5/2) ΔE = [(5/2) – (-5/2)] gⱼ μ_B B = 5 gⱼ μ_B B Substituting the values B = 0.5 T, μ_B = 5.788 × 10⁻⁵ eV/T, and gⱼ = 1.2: ΔE = 5 (1.2) (5.788 × 10⁻⁵ eV/T) * (0.5 T) ΔE = 6 (5.788 × 10⁻⁵ eV/T) (0.5 T) ΔE = 3 * (5.788 × 10⁻⁵ eV) = 1.7364 × 10⁻⁴ eV Therefore, the energy spacing between the mⱼ = +5/2 and mⱼ = -5/2 levels is 1.7364 × 10⁻⁴ eV .

Q4. Write down the basic assumptions of the Born-Oppenheimer approximation. Write the total wave function of a molecule under this approximation. Write the Hamiltonian for the electronic and nuclear motion. Discuss the physical significance of this approximation.

Ans. The Born-Oppenheimer approximation is a fundamental concept in molecular quantum mechanics that simplifies the determination of the wave function of a molecule. Basic Assumptions: The core assumption of the approximation is that nuclei are much heavier than electrons (the lightest nucleus, the proton, is ~1836 times more massive than an electron). Due to this large mass difference, the nuclei move much more slowly than the electrons. As a result, we can assume that:

1. The electrons respond instantaneously to the instantaneous position of the nuclei. In other words, when we solve for the electronic motion, we can treat the nuclei as being in a fixed configuration in space.
2. The nuclei move in an average potential created by the motion of all the electrons.

This approximation thus allows for the

separation of electronic and nuclear motion

.

Total Wave Function:

Under the Born-Oppenheimer approximation, the total wave function of the molecule, Ψ(

r

,

R

) (where

r

are the electronic coordinates and

R

are the nuclear coordinates), can be written as a product of the electronic and nuclear wave functions:

Ψ(r, R) ≈ ψₑₗ(r; R) χₙᵤ꜀(R)

Here, ψₑₗ(r; R) is the electronic wave function, which depends parametrically on the nuclear coordinates

R

. χₙᵤ꜀(R) is the nuclear wave function, describing the vibrational and rotational motion of the nuclei.

Hamiltonian:

The total Hamiltonian can be separated into two parts:

1.

Electronic Hamiltonian (Hₑₗ):

This describes the motion of the electrons in the field of the fixed nuclei. It includes the kinetic energy of the electrons (Tₑ), electron-electron repulsion (Vₑₑ), and electron-nuclear attraction (Vₑₙ). The nuclear kinetic energy (Tₙ) is neglected and the nucleus-nucleus repulsion (Vₙₙ) is a constant for this calculation.

Hₑₗ = Tₑ + Vₑₑ + Vₑₙ

The associated Schrödinger equation is: Hₑₗ ψₑₗ(r; R) = Eₑₗ(R) ψₑₗ(r; R). Here, the electronic energy Eₑₗ(R) depends on the nuclear configuration R.

2.

Nuclear Hamiltonian (Hₙᵤ꜀):

This describes the motion of the nuclei. It includes the nuclear kinetic energy (Tₙ) and an effective potential energy, which is the sum of the electronic energy Eₑₗ(R) and the nucleus-nucleus repulsion Vₙₙ(R).

Hₙᵤ꜀ = Tₙ + Eₑₗ(R) + Vₙₙ(R)

The term U(R) = Eₑₗ(R) + Vₙₙ(R) is called the

Potential Energy Surface (PES)

.

The associated Schrödinger equation is: Hₙᵤ꜀ χₙᵤ꜀(R) = E χₙᵤ꜀(R), where E is the total energy of the molecule.

Physical Significance:

The physical significance of the Born-Oppenheimer approximation is immense:

• It breaks down an extremely complex many-body problem into two simpler, separate problems: one electronic and one nuclear.
• It introduces the concept of the Potential Energy Surface (PES) , which is central to the understanding of molecular structure and chemistry. The PES allows us to predict the equilibrium geometry of a molecule, bond energies, vibrational frequencies, and activation barriers for chemical reactions.
• It enables us to understand and interpret the vibrational and rotational spectra of molecules, as these arise from the nuclear motion that takes place on the PES.

In essence, this approximation is the foundation of modern computational chemistry and molecular spectroscopy.

Q5. (a) Obtain the normalization constant for the eigenfunction ψ₋ = N₋(1sₐ – 1s♭), where 1sₐ and 1s♭ are the hydrogen 1s-type orbitals centered on atom ‘a’ and atom ‘b’ respectively. Express the normalization constant in terms of the overlap integral Sₐ♭. (b) Consider a molecular electronic configuration πg⁴ for a two-electron system. Determine the term symbols associated with the configuration.

Ans. (a) The normalization condition for a wave function ψ is <ψ|ψ> = ∫ψ*ψ dτ = 1. For the given antibonding molecular orbital: ψ₋ = N₋(1sₐ – 1s♭) Assuming the 1s orbitals are real functions, ψ* = ψ. The normalization condition becomes: ∫[N₋(1sₐ – 1s♭)]² dτ = 1 N₋² ∫(1sₐ² – 2(1sₐ)(1s♭) + 1s♭²) dτ = 1 Splitting the integral into terms: N₋² [∫1sₐ² dτ + ∫1s♭² dτ – 2∫(1sₐ)(1s♭) dτ] = 1 Now we identify each term:

• ∫1sₐ² dτ = <1sₐ|1sₐ> = 1 (since the atomic orbital 1sₐ is normalized).
• ∫1s♭² dτ = <1s♭|1s♭> = 1 (since the atomic orbital 1s♭ is normalized).
• ∫(1sₐ)(1s♭) dτ = <1sₐ|1s♭> = Sₐ♭ (by definition, this is the overlap integral between the atomic orbitals).

Substituting these values back into the equation:

N₋² [1 + 1 – 2Sₐ♭] = 1

N₋² [2(1 – Sₐ♭)] = 1

N₋² = 1 / [2(1 – Sₐ♭)]

Therefore, the normalization constant N₋ is:

N₋ = 1 / √[2(1 – Sₐ♭)]

(b)

Note: The question mentions a πg⁴ configuration, which refers to 4 electrons. However, it is described as a “two-electron system.” This is a typo. We will proceed assuming the question means a

(πg)² configuration

in a homonuclear diatomic molecule.

For a (πg)² configuration, we have two electrons in the πg orbital. A π orbital has azimuthal angular momentum projection quantum numbers mₗ = ±1.

We calculate the total orbital angular momentum projection, Mₗ = mₗ₁ + mₗ₂, and total spin angular momentum projection, Mₛ = mₛ₁ + mₛ₂.

To apply the Pauli Exclusion Principle, we consider the possible microstates (mₗ₁, mₛ₁; mₗ₂, mₛ₂).

Possible Mₗ values:

• mₗ₁=+1, mₗ₂=+1 → Mₗ = +2
• mₗ₁=-1, mₗ₂=-1 → Mₗ = -2
• mₗ₁=+1, mₗ₂=-1 → Mₗ = 0

So, we have states with a maximum value of |Mₗ|=2, indicating a

Δ state (L=2)

, and states with Mₗ=0, indicating

Σ states (L=0)

.

Now we consider spin and the Pauli principle:

1.

For Mₗ = ±2 (Δ state):

Both electrons are in the same spatial orbital (either mₗ=+1 or mₗ=-1). By the Pauli principle, their spins must be opposed (one mₛ=+1/2, the other -1/2). Therefore, Mₛ = 0, which implies a total spin S=0 (singlet). This gives us a

¹Δ

term.

2.

For Mₗ = 0 (Σ states):

One electron is in mₗ=+1 and the other is in mₗ=-1. Since they are in different spatial orbitals, their spins can be either parallel (S=1, triplet) or anti-parallel (S=0, singlet).

• S=1 (Triplet): Mₛ = +1, 0, -1. This gives rise to a ³Σ term.
• S=0 (Singlet): Mₛ = 0. This gives rise to a ¹Σ term.

Now we determine the symmetry properties:

• Parity (g/u): Since both electrons are in g (gerade) orbitals, the total parity is g × g = g . All terms will be g.
• Reflection symmetry (+/-) for Σ states:
  • For the ¹Σ state , the spin wavefunction is antisymmetric, so the spatial wavefunction must be symmetric. The symmetric spatial combination is symmetric (+1) under reflection in a plane containing the internuclear axis. Thus, it is a ¹Σg⁺ term.
  • For the ³Σ state , the spin wavefunction is symmetric, so the spatial wavefunction must be antisymmetric. The antisymmetric spatial combination is antisymmetric (-1) under reflection. Thus, it is a ³Σg⁻ term.

Therefore, the term symbols associated with the (πg)² configuration are:

³Σg⁻, ¹Δg, ¹Σg⁺

.

Q6. The vibrational spectrum of HBr shows a very intense absorption at 2649 cm⁻¹ and a weaker absorption at 5188 cm⁻¹. Determine the equilibrium oscillation frequency (ωₑ), the anharmonicity constant (xₑ), the zero-point energy, and the force constant (k) of the molecule.

Ans. The vibrational energy levels (in cm⁻¹) of an anharmonic oscillator are given by: G(v) = ωₑ(v + 1/2) – ωₑxₑ(v + 1/2)² where ωₑ is the equilibrium oscillation frequency, xₑ is the anharmonicity constant, and v is the vibrational quantum number. The intense absorption corresponds to the fundamental transition (v=0 → v=1), and the weaker absorption corresponds to the first overtone (v=0 → v=2). (The value of 5188 cm⁻¹ is used, as 588 cm⁻¹ from the image text is a clear typographical error.) 1. Fundamental Transition (ν̃₁): ν̃₁ = G(1) – G(0) = [ωₑ(3/2) – ωₑxₑ(9/4)] – [ωₑ(1/2) – ωₑxₑ(1/4)] ν̃₁ = ωₑ – 2ωₑxₑ = 2649 cm⁻¹ — (Eq. 1) 2. First Overtone (ν̃₂): ν̃₂ = G(2) – G(0) = [ωₑ(5/2) – ωₑxₑ(25/4)] – [ωₑ(1/2) – ωₑxₑ(1/4)] ν̃₂ = 2ωₑ – 6ωₑxₑ = 5188 cm⁻¹ — (Eq. 2) We now have a system of two equations for ωₑ and ωₑxₑ. Multiply Eq. 1 by 3: 3ωₑ – 6ωₑxₑ = 3 × 2649 = 7947 cm⁻¹ — (Eq. 3) Subtract Eq. 2 from Eq. 3: (3ωₑ – 6ωₑxₑ) – (2ωₑ – 6ωₑxₑ) = 7947 – 5188 ωₑ = 2759 cm⁻¹ Now, substitute the value of ωₑ into Eq. 1: 2759 – 2ωₑxₑ = 2649 2ωₑxₑ = 2759 – 2649 = 110 ωₑxₑ = 55 cm⁻¹ The anharmonicity constant xₑ is: xₑ = (ωₑxₑ) / ωₑ = 55 / 2759 ≈ 0.0199 3. Zero-Point Energy (ZPE): The ZPE is the energy of the v=0 level: G(0) = ωₑ(1/2) – ωₑxₑ(1/4) G(0) = (2759 / 2) – (55 / 4) = 1379.5 – 13.75 = 1365.75 cm⁻¹ 4. Force Constant (k): The force constant is related to ωₑ: ωₑ (in s⁻¹) = c ωₑ (in cm⁻¹) = (1 / 2π) √(k / μ) k = (2πc ωₑ_cm⁻¹)² μ Calculate the reduced mass μ of HBr: m_H = 1.00784 amu, m_Br = 79.904 amu μ = (m_H m_Br) / (m_H + m_Br) = (1.00784 79.904) / (1.00784 + 79.904) = 80.536 / 80.91184 ≈ 0.99535 amu μ = 0.99535 amu * (1.66054 × 10⁻²⁷ kg/amu) ≈ 1.6528 × 10⁻²⁷ kg Now calculate k: k = [2π (3 × 10¹⁰ cm/s) (2759 cm⁻¹)]² * (1.6528 × 10⁻²⁷ kg) k = [5.199 × 10¹⁴ s⁻¹]² * (1.6528 × 10⁻²⁷ kg) k = (2.703 × 10²⁹ s⁻²) * (1.6528 × 10⁻²⁷ kg) k ≈ 412.3 N/m Summary:

• Equilibrium oscillation frequency (ωₑ) = 2759 cm⁻¹
• Anharmonicity constant (xₑ) = 0.0199 (or ωₑxₑ = 55 cm⁻¹)
• Zero-point energy = 1365.75 cm⁻¹
• Force constant (k) = 412.3 N/m

Q7. The rotational constant for the excited electronic state (B’) of a diatomic molecule is 25% smaller than the rotational constant of its ground electronic state (B”). Determine the values and the successive separation for the first three lines of the P-branch and R-branch, where v’ = 1, v” = 0, B” = 1.8 cm⁻¹ and ΔG(v’=1,v”=0) = 5000 cm⁻¹.

Ans. In this problem, we analyze the rovibrational spectrum of a diatomic molecule. Given parameters:

• Band origin, ν̃₀ = ΔG(v’=1,v”=0) = 5000 cm⁻¹
• Rotational constant of the ground state, B” = 1.8 cm⁻¹
• Rotational constant of the excited state, B’, is 25% smaller than B”: B’ = B” – 0.25 B” = 0.75 B” = 0.75 * 1.8 cm⁻¹ = 1.35 cm⁻¹

We will calculate the positions of the lines in the R-branch (ΔJ = J’ – J” = +1) and P-branch (ΔJ = J’ – J” = -1).

The formula for the lines in the R-branch is:

ν̃_R(J”) = ν̃₀ + (B’ + B”)(J” + 1) + (B’ – B”)(J” + 1)² (for J” = 0, 1, 2, …)

The formula for the lines in the P-branch is:

ν̃_P(J”) = ν̃₀ – (B’ + B”)J” + (B’ – B”)J”² (for J” = 1, 2, 3, …)

Calculate the constants:

B’ + B” = 1.35 + 1.8 =

3.15 cm⁻¹

B’ – B” = 1.35 – 1.8 =

-0.45 cm⁻¹

R-branch:

• First line (J” = 0): ν̃_R(0) = 5000 + (3.15)(1) + (-0.45)(1)² = 5000 + 3.15 – 0.45 = 5002.7 cm⁻¹
• Second line (J” = 1): ν̃_R(1) = 5000 + (3.15)(2) + (-0.45)(2)² = 5000 + 6.30 – 1.80 = 5004.5 cm⁻¹
• Third line (J” = 2): ν̃_R(2) = 5000 + (3.15)(3) + (-0.45)(3)² = 5000 + 9.45 – 4.05 = 5005.4 cm⁻¹

Successive separation in the R-branch:

• ν̃_R(1) – ν̃_R(0) = 5004.5 – 5002.7 = 1.8 cm⁻¹
• ν̃_R(2) – ν̃_R(1) = 5005.4 – 5004.5 = 0.9 cm⁻¹

The separation is decreasing, indicating the formation of an R-branch band head, as expected for B’ < B”.

P-branch:

• First line (J” = 1): ν̃_P(1) = 5000 – (3.15)(1) + (-0.45)(1)² = 5000 – 3.15 – 0.45 = 4996.4 cm⁻¹
• Second line (J” = 2): ν̃_P(2) = 5000 – (3.15)(2) + (-0.45)(2)² = 5000 – 6.30 – 1.80 = 4991.9 cm⁻¹
• Third line (J” = 3): ν̃_P(3) = 5000 – (3.15)(3) + (-0.45)(3)² = 5000 – 9.45 – 4.05 = 4986.5 cm⁻¹

Successive separation in the P-branch:

• ν̃_P(1) – ν̃_P(2) = 4996.4 – 4991.9 = 4.5 cm⁻¹
• ν̃_P(2) – ν̃_P(3) = 4991.9 – 4986.5 = 5.4 cm⁻¹

The separation is increasing, which is typical for the P-branch.

Q8. (a) The selenium soft X-ray laser operates at a wavelength of 20.6 nm in ⁸⁰Se²⁴⁺ ions. These ions are produced at a plasma temperature of approximately 10⁷ K. How large is the Doppler width of this laser compared with the Doppler width in a He-Ne laser operating at 632.8 nm and 100°C ? Here the atomic mass of selenium is 80 a.m.u. and the atomic mass of Neon is 20 a.m.u. (b) Give the main difference between a Ruby laser and a free electron laser.

Ans. (a) Doppler broadening is due to the thermal motion of the emitting atoms or ions. The formula for the Doppler width in wavelength (Δλ_D) is: Δλ_D = (2λ₀/c) * √[(2k_B T ln2) / M] where λ₀ is the central wavelength, c is the speed of light, k_B is the Boltzmann constant, T is the temperature in Kelvin, and M is the mass of the emitting particle. We want to calculate the ratio of the Doppler widths of the two lasers: (Δλ_D)_Se / (Δλ_D)_Ne. Ratio = [(2λ_Se/c) √[(2k_B T_Se ln2) / M_Se]] / [(2λ_Ne/c) √[(2k_B T_Ne ln2) / M_Ne]] The constants (2, c, k_B, ln2) cancel out, simplifying this to: Ratio = (λ_Se / λ_Ne) * √[(T_Se / M_Se) / (T_Ne / M_Ne)] Ratio = (λ_Se / λ_Ne) √[(T_Se / T_Ne) (M_Ne / M_Se)] Substitute the given values: Selenium (Se) laser:

• λ_Se = 20.6 nm
• T_Se = 10⁷ K
• M_Se = 80 a.m.u.

Helium-Neon (He-Ne) laser:

(we use Neon as the emitting species)

• λ_Ne = 632.8 nm
• T_Ne = 100°C = 100 + 273.15 = 373.15 K
• M_Ne = 20 a.m.u.

Now calculate the ratio:

Ratio = (20.6 / 632.8)

√[(10⁷ / 373.15)

(20 / 80)]

Ratio = (0.03255)

√[(26799)

(0.25)]

Ratio = 0.03255 * √[6699.75]

Ratio = 0.03255 * 81.85

Ratio ≈

2.66

Therefore,

the Doppler width of the selenium soft X-ray laser is about 2.66 times larger than that of the He-Ne laser.

The extremely high plasma temperature is the main contributor to this large broadening, even though its wavelength is much shorter.

(b) The main differences between a Ruby laser and a Free Electron Laser (FEL) are:

1. Gain Medium:
   • Ruby Laser: Uses a solid-state gain medium, which is chromium ions (Cr³⁺) doped into an aluminum oxide (Al₂O₃) crystal. The lasing action occurs between the fixed, discrete electronic energy levels of these ions. It is a three-level laser system.
   • Free Electron Laser (FEL): Uses a beam of relativistic electrons passing through a periodic magnetic structure (called a ‘wiggler’ or ‘undulator’) as its gain medium. It does not rely on bound energy levels of atoms or molecules.
2. Tunability:
   • Ruby Laser: Has a fixed output wavelength (around 694.3 nm), determined by the energy levels of the Cr³⁺ ions.
   • Free Electron Laser (FEL): Are highly tunable . The output wavelength can be adjusted over a wide range, from microwaves to X-rays, by changing the energy of the electron beam or the magnetic field strength and periodicity of the wiggler.
3. Operating Principle:
   • Ruby Laser: Works on the principle of population inversion and stimulated emission between energy levels.
   • FEL: Works on the principle of synchrotron radiation. As electrons oscillate through the wiggler, they emit light. This light interacts with the electrons, causing them to gather into bunches which then radiate coherently, amplifying the laser light.

In short, the key difference lies in the gain medium (crystal vs. electron beam) and the resulting tunability (fixed vs. widely tunable).