This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Spectrum for Hydrogen Like Atoms”.
1. What is the value of the Rydberg constant R in inverse centimeters (for the hydrogen atom)?
a) 1.097 × 105cm-1
b) 9.381 × 1015cm-1
c) 5.322 × 106cm-1
d) 2.347 × 109cm-1
View Answer
Explanation: Rydberg Constant is a congregation of all other constants in the energy levels equation for the hydrogen atom. For the hydrogen atom, it is given by R = \(\frac{\mu e^4}{4\pi c(4\pi \varepsilon_0)^2 \hbar^3 n^2}\) = 1.097 × 107m-1 = 1.097 × 105cm-1.
2. What is the ionization energy of hydrogen in eV?
a) 122.454 eV
b) 217.696 eV
c) 54.24 eV
d) 13.606 eV
View Answer
Explanation: Ionization energy is the energy required to eject an electron from the outermost energy level of an atom. It is given by the expression Ei = (13.606 × Z2)eV = 13.606 × 12 = 13.606eV for hydrogen.
3. What is the ionization energy of Be3+ in eV?
a) 122.454 eV
b) 217.696 eV
c) 54.24 eV
d) 13.606 eV
View Answer
Explanation: Ionization energy is the energy required to eject an electron from the outermost energy level of an atom. It is given by the expression Ei = (13.606 × Z2)eV = 13.606 × 42 = 217.696eV for a beryllium 3+ ion.
4. What is the total degeneracy of a hydrogen like atom in terms of quantum numbers n, l, and m?
a) n2
b) n × l × m
c) \(\frac{4\pi \varepsilon_0}{n × l × m}\)
d) \(\frac{l^2}{2m}\)
View Answer
Explanation: One can show the total degeneracy gtotal of a hydrogen like atom from the table below (only shown for first three principle quantum numbers):
| Orbital | n | l | m | g | gtotal |
|---|---|---|---|---|---|
| 1s | 1 | 0 | 0 | 1 | 1 |
| 2s | 2 | 0 | 0 | 1 | |
| 2p | 1 | 0, ±1 | 3 | 4 | |
| 3s | 3 | 0 | 0 | 1 | |
| 3p | 1 | 0, ±1 | 3 | ||
| 3d | 2 | 0, ±1 ±2 | 5 | 9 |
Thus, the total degeneracy always adds up to be n2. This can also be proved by mathematical induction.
5. The Schrodinger equation for a hydrogen like atom can be solved exactly and is given by φn,l,m(r,θ,Φ) = Nnlρl\(e^{-\frac{ρ}{2}}L_{n+l}^{2l+1} (\rho)Y_l^m\)(θ,Φ), where Nnl is a normalization constant. What does \(Y_l^m\) (θ,Φ) signify?
a) Spherical harmonic function
b) Cylindrical harmonic function
c) Fourier transformed function
d) Gaussian function
View Answer
Explanation: Being a function of &theta, Φ,\(Y_l^m\) (θ,Φ) must be a spherical function as radial angle and azimuthal angle only appear in spherical coordinates. These are complex functions depending on quantum number n, l, and m. Spherical harmonics show a complete set of orthogonal functions on the surface of a sphere.
6. What are the allowed values of angular momentum quantum number (l) in terms of given n (principal quantum number)?
a) l = \(\frac{1}{2n}\)
b) l = \(\frac{1}{n^2}\)
c) l = n2
d) l = n – 1
View Answer
Explanation: For a given value of n, there are always one less allowed number of total orbital angular momentum quantum numbers. L describes the shape of the orbital (s, p, d, f) in space and forms an important part of how complex atomic functions are depicted mathematically. A complete specification of n, l, and m describes a particular electron in an atom.
7. What are the allowed values of magnetic quantum number (m) in terms of given (angular momentum quantum number)?
a) m = ±l
b) m = -l,…0…..,+l
c) m = \(\frac{\pm l}{2\pi}\)
d) m = l – 1
View Answer
Explanation: Values of m range in the interval of negative to positive l. The magnetic quantum number determines the z-component of angular momentum. It also tells one more about the spin nature of an electron residing in an orbital.
8. What is the force constant k of a chlorine molecule in SI units if it is treated as a harmonic oscillator? Vibrational frequency of chlorine is 560 cm-1.
a) 9201 Nm-1
b) 323 Nm-1
c) 219 Nm-1
d) 2920 Nm-1
View Answer
Explanation: Reduced mass of chlorine: μ = \(\frac{35×35}{35+35}\) × 1.66 × 10-27 kg = 2.905 × 10-26 kg. Force constant k is given by k = ω2 μ = (2πv)2 μ = (2 × π × 3 × 1010cms-1 × 560 cm-1)2 × 2.905 × 10-26kg ∼ 323 Nm-1
9. What is the force constant k of a hydrogen molecule (amu = 1) in SI units if it is treated as a harmonic oscillator? Vibrational frequency of hydrogen is 4401 cm-1.
a) 571 Nm-1
b) 393 Nm-1
c) 173 Nm-1
d) 3910 Nm-1
View Answer
Explanation: Reduced mass of hydrogen: μ = \(\frac{1×1}{1+1}\) × 1.66 × 10-27kg = 8.3 × 10-28kg. Force constant k is given by k = ω2μ = (2πv)2 μ = (2 × π × 3 × 1010cms-1 × 4401 cm-1)2 × 8.3 × 10-28kg ∼ 571 Nm-1.
10. What is the force constant k of a carbon monoxide molecule in SI units (amu = 12 & 16) if it is treated as a harmonic oscillator? Vibrational frequency of carbon monoxide is 2170 cm-1.
a) 323 Nm-1
b) 4921 Nm-1
c) 3013 Nm-1
d) 1900 Nm-1
View Answer
Explanation: Reduced mass of carbon monoxide: μ = \(\frac{16×12}{12+16}\) × 1.66 × 10-27kg = 1.14 × 10-26kg. Force constant k is given by k = ω2 μ = (2πv)2 μ = (2 × π × 3 × 1010cms-1 × 2170 cm-1)2 × 1.14 × 10-26kg ∼ 1900 Nm-1.
11. What is the value of standard deviation in internuclear distance for a carbon monoxide molecule (amu = 12 & 16) in picometers? Vibrational frequency is 2170 inverse centimeters.
a) 3.4 pm
b) 201.3 pm
c) 49.1 pm
d) 1.1 pm
View Answer
Explanation: Reduced mass of carbon monoxide: μ = \(\frac{16×12}{12+16}\) × 1.66 × 10-27kg = 1.14 × 10-26kg. Force constant k is given by k = ω2 μ = (2πv)2 μ = (2 × π × 3 × 1010cms-1 × 2170 cm-1)2 × 1.14 × 10-26kg ∼ 1900 Nm-1. Standard deviation Δx = \((\frac{1}{2\alpha})^{1/2}\) = \((\frac{\hbar^2}{4k\mu})^{1/4}\) = \((\frac{(1.05×10^{-34})^2}{4×1900×1.14×10^{-26}})^{1/4}\) ≈ 3.4 pm
12. What is the value of standard deviation in momentum in vibrational motion for a carbon monoxide molecule (amu = 12 & 16) in SI units? Vibrational frequency is 2170 inverse centimeters.
a) 8.29 × 10-20kgms-1
b) 9.12 × 10-25kgms-1
c) 1.56 × 10-23kgms-1
d) 4.32 × 10-29kgms-1
View Answer
Explanation: μ = \(\frac{16×12}{12+16}\) × 1.66 × 10-27kg = 1.14 × 10-26kg. Force constant k is given by k = ω2 μ = (2πv)2 μ = (2 × π × 3 × 1010cms-1 × 2170 cm-1)2 × 1.14 × 10-26kg ∼ 1900 Nm-1. Standard deviation Δx = \((\frac{\alpha\hbar^2}{2})^{1/2} = (\frac{k\mu \hbar^2}{4})^{1/4} = (\frac{1900×1.14×10^{-26}×(1.05×10^{-34})^2}{4})^{1/4}\) ≈ 1.56 × 10-23kgms-1.
Sanfoundry Global Education & Learning Series – Physical Chemistry.
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