This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Orbital Angular Momentum of Hydrogen Like Atoms”.
1. What quantum number does the energy of a system depend on in the presence of a magnetic field?
a) Doesn’t depend on any quantum number
b) n
c) l
d) m
View Answer
Explanation: In the presence of a magnetic field, the energy depends on the magnetic quantum number m. This tells one about electron spin, and this can be oriented differently in the presence of a magnetic field. The magnetic field removes the 2l + 1 degeneracy with respect to m.
2. The Bohr Magneton is given by μB = \(\frac{e\hbar}{2m_e}\). What is the numerical value of this Bohr Magneton?
a) 9.27 × 10-24JT-1
b) 4.13 × 10-15JT-1
c) 2.35 × 10-20JT-1
d) 6.24 × 10-10JT-1
View Answer
Explanation: μB = \(\frac{e\hbar}{2m_e} = \frac{(1.602×10^{-19}C)(1.055×10^{-34}Js}{2(9.109×10^{-31}kg})\) = 9.274 × 10-24JT-1.
3. What is the Zeeman effect?
a) Splitting of spectral lines due to a magnetic field
b) Change in Hamiltonian due to a magnetic field
c) Induced magnetic dipole due to an electric field
d) Induced electric dipole due to a magnetic field
View Answer
Explanation: Energy level splitting in the presence of a magnetic field is studied by measuring magnetic susceptibility by electron paramagnetic resonance. The splitting can be contributed due to intrinsic (spin) magnetic moment or extrinsic magnetic moment.
4. For a hydrogen atom with a 3d electron, what are the possible values of m?
a) m = 0
b) m = 0, ±1, ±2
c) m = ±1
d) m = ±1, ±2
View Answer
Explanation: Since the electron is in a d-orbital, its angular momentum quantum number is l = 2 and its magnetic quantum number can range from m = -2 to m = +2. Its spectral line can thus split into 5 separate lines due to the magnetic field for the given 5 values of m.
5. Consider a Hydrogen atom in the superposition state φ = \(\frac{1}{\sqrt 2}\) |2s > + \(\frac{1}{2}\)|3p-1 > + \(\frac{1}{2}\)|3p0 >. By what number should this be divided by for normalization?
a) 1
b) √2
c) 4
d) 2
View Answer
Explanation: The wavefunction is already normalized: ∑|Cn|2 = \(\frac{1}{2} + \frac{1}{4} + \frac{1}{4}\) = 1. Hence the function only must be divided by 1 for normalization.
6. Consider a Hydrogen atom in the superposition state φ = \(\frac{1}{\sqrt 2}\) |2s > + \(\frac{1}{2}\)|3p-1 > + \(\frac{1}{2}\)|3p0 >. What is the expectation value of this superposition state?
a) –\(\frac{1}{2}\)RH
b) –\(\frac{77}{391}\)RH
c) –\(\frac{43}{576}\)RH
d) –\(\frac{13}{72}\)RH
View Answer
Explanation: < E > = ∑|Cn|2 \((-\frac{R_H}{n^2}) = \frac{1}{2}(-\frac{R_H}{4}) + \frac{1}{4}(-\frac{R_H}{9}) + \frac{1}{4}(-\frac{R_H}{9}) = -\frac{13}{72}\)RH.
7. Consider a Hydrogen atom in the superposition state φ = \(\frac{1}{\sqrt 2}\) |2s > + \(\frac{1}{2}\)|3p-1 > + \(\frac{1}{2}\)|3p0 >. What is one possible outcome of L2 and with what probability?
a) 4ℏ2, probability: 0.75
b) 4ℏ2, probability: 0.5
c) 0, probability: 0.75
d) 0, probability: 0.5
View Answer
Explanation: For the 2s and 3p zero state, the l value is 0 as well. Hence, L2 = ℏ2l(l + 1) = 0. The probability of this is half because: \((\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{2}\).
8. Consider a Hydrogen atom in the superposition state φ = \(\frac{1}{\sqrt 2}\) |2s > + \(\frac{1}{2}\)|3p-1 > + \(\frac{1}{2}\)|3p0 >. What is one possible outcome of L2 and with what probability?
a) 4ℏ2, probability: 0.75
b) 2ℏ2, probability: 0.5
c) 0, probability: 0.75
d) 0, probability: 0.25
View Answer
Explanation: For the 3p-1 state, the l value is 1. Hence, L2 = ℏ2l(l + 1) = 2ℏ2. Probability = \((\frac{1}{\sqrt 2})^2 = \frac{1}{2}\)
9. Consider a Hydrogen atom in the superposition state φ = \(\frac{1}{\sqrt 2}\) |2s > + \(\frac{1}{2}\)|3p-1 > + \(\frac{1}{2}\)|3p0 >. What is the expectation value of the z-component of angular momentum?
a) 0
b) ℏ/2
c) -ℏ/4
d) -ℏ/2
View Answer
Explanation: < Lz > = ∑|Cn|2 ℏm = \(\frac{1}{4}\)(-ℏ) = –\(\frac{\hbar}{4}\).
10. How many radial and angular nodes are there for electrons in the 3d orbital?
a) Angular nodes: 2, Radial nodes: 0
b) Angular nodes: 1, Radial nodes: 1
c) Angular nodes: 0, Radial nodes: 2
d) Angular nodes: 2, Radial nodes: 1
View Answer
Explanation: # Angular nodes is equal to the angular momentum quantum number = l = 2. # radial nodes = n-l-1 = 3-2-1 = 0. These nodes will be seen in the wavefunction plot as well. Nodes help determine various quantum numbers, and hence the orbital of the given wavefunction.
11. How many radial and angular nodes are there for electrons in the 3p orbital?
a) Angular nodes: 2, Radial nodes: 0
b) Angular nodes: 1, Radial nodes: 1
c) Angular nodes: 0, Radial nodes: 2
d) Angular nodes: 2, Radial nodes: 1
View Answer
Explanation: The number of radial nodes is given by n-l-1, whereas the number of angular nodes is just l. A 3p orbital has n = 3 and l = 1, hence plugging these into the previous equations gives the number of angular and radial nodes as 1.
12. What is the difference in energy between two spin angular momentum states of a hydrogen atom in a 1s orbital, experiencing a magnetic field of 1 Tesla?
a) 6.31 × 10-20J
b) 1.59 × 10-15J
c) 3.13 × 10-25J
d) 1.86 × 10-23J
View Answer
Explanation: ΔS = 1; ΔE = geμBΔSB = 2.0023 × 9.274 × 10-24 ≈ 1.86 × 10-23J. Hence radiation of this energy is emitted and there is a split in spectral lines for these two spin angular momentum states.
13. What is the type of radiation emitted between two spin angular momentum states of a hydrogen atom in a 1s orbital, experiencing a magnetic field of 1 Tesla?
a) Gamma
b) Infrared
c) Radiowave
d) Microwave
View Answer
Explanation: ΔS = 1; ΔE = geμBΔSB = 2.0023 × 9.274 × 10-24 ≈ 1.86 × 10-23J. λ = \(\frac{hc}{\Delta E} = \frac{6.626 ×10^{-34}×3×10^{8}}{1.86×10^{-23}}\) ∼ 1.07 × 10-2m. This wavelength falls in the microwave radiation range. Hence radiation of this energy is emitted and there is a split in spectral lines for these two spin angular momentum states.
14. When wavefunctions are a superposition of real and complex parts, how is the overall real wavefunction written?
a) As a linear combination of the real and complex wavefunctions such that the complex part is eliminated
b) The overall real wavefunction only consists of the real part of the superimposed wavefunction
c) The wavefunction is written as a superimposed function of real and complex parts
d) Complex terms are written as exponentials using Euler’s formula
View Answer
Explanation: If there are two functions, φ2 p+1 = \(\frac{1}{8\sqrt{\pi}}(\frac{Z}{a})^5 e^{-\frac{Zr}{2a}}\)r sin θ(cos φ + sin φ), and φ2 p-1 = \(\frac{1}{8\sqrt{\pi}}(\frac{Z}{a})^5 e^{-\frac{Zr}{2a}}\)r sin θ(cos φ – sin φ), a real linear superposition function will be written as φ = \(\frac{1}{\sqrt 2}\)(φ2 p+1 + φ2 p-1)
Sanfoundry Global Education & Learning Series – Physical Chemistry.
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