This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Spectrum for Hydrogen Like Atoms – Set 2”.
1. What is an algebraic expression for energy required to move an electron from J = 1 to J = 2 in an HCl molecule in terms of reduced mass and equilibrium radius?
a) ΔE = \(\mu R_e^2\)
b) ΔE = \(\frac{2\pi \varepsilon_0}{\mu R_e^2}\)
c) ΔE = \(\frac{2\hbar^2}{\mu R_e^2}\)
d) ΔE = \(\frac{\mu R_e^2}{2\hbar^2}\)
View Answer
Explanation: Energy levels in terms of J is given by EJ = \(\frac{\hbar^2}{2I}\)J(J+1). Replacing I with \(\mu R_e^2\) and plugging in values of J gives EJ = \(\frac{\hbar^2}{2\mu R_e^2}\)(2.3 – 1.2) = \(\frac{2\hbar^2}{\mu R_e^2}\)
2. What is the frequency of radiation required to take an HCl molecule from J = 1 to J = 2?
a) 1.27 × 1012s-1
b) 2.24 × 109s-1
c) 8.19 × 1010s-1
d) 6.42 × 107s-1
View Answer
Explanation: Energy levels in terms of J is given by EJ = \(\frac{\hbar^2}{2I}\)J(J+1). Replacing I with \(\mu R_e^2\) and plugging in values of J gives EJ = \(\frac{\hbar^2}{2\mu R_e^2}\)(2.3 – 1.2) = \(\frac{2\hbar^2}{\mu R_e^2}\). Frequency v = \(\frac{\Delta E}{h} = \frac{h}{2\pi^2 I} = \frac{6.626×10^{-34}}{2×\pi^2×2.64×10^{-47}}\) ≈ 1.27 × 1012s-1
3. What is the wavenumber of radiation required to take an HCl molecule from J = 1 to J = 2?
a) 42.3 cm-1
b) 29.1 cm-1
c) 82.3 cm-1
d) 39.6 cm-1
View Answer
Explanation: Energy levels in terms of J is given by EJ = \(\frac{\hbar^2}{2I}\)J(J+1). Replacing I with \(\mu R_e^2\) and plugging in values of J gives EJ = \(\frac{\hbar^2}{2\mu R_e^2}\)(2.3 – 1.2) = \(\frac{2\hbar^2}{\mu R_e^2}\). Frequency v = \(\frac{\Delta E}{h} = \frac{h}{2\pi^2 I} = \frac{6.626×10^{-34}}{2×\pi^2×2.64×10^{-47}}\) ≈ 1.27 × 1012s-1. Wavenumber is given by v∼ = \(\frac{v}{c} = \frac{1.27×10^{12} s^{-1}}{3×10^{10} cms^{-1}}\) = 42.3 cm-1
4. What is the reduced mass of a carbon monoxide molecule in kilograms?
a) 2.53 × 10-23kg
b) 1.14 × 10-26kg
c) 8.01 × 10-10kg
d) 3.14 × 10-20kg
View Answer
Explanation: μCO = \(\frac{m_C m_o}{m_c+m_o} = \frac{12.16}{12+16}\) × 1.66 × 10-27kg ≈ 1.14 × 10-26kg
5. What is the moment of inertia of a carbon monoxide molecule? Re = 112.8 pm
a) 1.45 × 10-46kgm2
b) 5.24 × 10-23kgm2
c) 9.34 × 103kgm2
d) 8.53 × 10-7kgm2
View Answer
Explanation: Reduced mass μCO = \(\frac{m_C m_o}{m_c+m_o} = \frac{12.16}{12+16}\) × 1.66 × 10-27kg ≈ 1.14 × 10-26kg. Moment of inertia = \(\mu R_e^2\) = 1.14 × 10−26 × (112.8 × 10−12)2 = 1.45 × 10−46kgm2
6. What is the algebraic expression for change in energy when a photon is absorbed from l = 1 to l = 0?
a) ΔE = \(\frac{\hbar^2}{I}\)
b) ΔE = \(\frac{\hbar^2I}{I}\)
c) ΔE = 2πIħ2
d) ΔE = \(\frac{\hbar^2}{2\pi I}\)
View Answer
Explanation: Energy levels are given by E = \(\frac{l(l+1)ℏ^2}{2I}\). From l = 1 to l = 0 the energy absorbed is ΔE = E1 – E0 = \(\frac{\hbar^2}{2I}\) (1.2 – 0.1) = \(\frac{\hbar^2}{I}\)
7. What Is the ionization energy of a hydrogen atom (in eV) given the energy of formations for hydrogen and hydrogen cation are 216.035 and 1528.085 kJ/mole respectively?
a) 3.424 eV
b) 393.429 eV
c) 13.606 eV
d) 294.319 eV
View Answer
Explanation: Ionization energy is the amount of energy required to eject an electron from the outermost shell of an atom. Since hydrogen cation has no electron, the ionization energy E will just be the difference between the energy of formations: E = \(\frac{(1528.085-216.035 \frac{kJ}{mole})}{96.485 \frac{kJ}{mole.eV}}\) = 13.606 eV.
8. What is the most probable distance for the 1s orbital from proton to electron for a hydrogen like atom in terms of Bohr’s radius and atomic number? Wavefunction for this orbital is φ1s = (\(\frac{Z^3}{\pi a_0^3})^{1/2} e^{-\frac{Zr}{a_0}}\)
a) \(\frac{a_0}{Z}\)
b) \(\frac{a_0.Z}{2\pi}\)
c) \(\frac{a_0}{Z}\).2π
d) a0Z
View Answer
Explanation: Radial probability density dP = |φ1S|24πr2 dr = \(\frac{4Z^3}{a_0^3} r^2 e^{-\frac{2Zr}{a_0}}\)dr. For most probable distance, the derivative is taken and equated to 0. This gives 2r\(e^{-\frac{2Zr}{a_0}} (1 – \frac{Zr}{a_0}\)) = 0 . Solving this equation gives r = \(\frac{a_0}{Z}\)
9. Expected value for radial distance from nucleus in terms of quantum numbers n and l is given by < r >nl = \(\frac{n^2 a_0}{Z} \Big\{1 + \frac{1}{2} \Big[1 – \frac{l(l+1)}{n^2}\Big]\Big\}\). What is the average distance from an orbital electron to the nucleus for a 2s orbital in a hydrogen atom in picometers?
a) 88.2 pm
b) 317.5 pm
c) 264.6 pm
d) 105.8 pm
View Answer
Explanation: For a 2s hydrogen orbital: n = 2, l = 0, Z = 1. Plugging these into the given equation yields < r >nl = \(\frac{2^2 a_0}{Z}\)(1 + \(\frac{1}{2}\)) = 6a0 ≈ 317.5 pm
10. Expected value for radial distance from nucleus in terms of quantum numbers n and l is given by < r >nl = \(\frac{n^2 a_0}{Z} \Big\{1 + \frac{1}{2} \Big[1 – \frac{l(l+1)}{n^2}\Big]\Big\}\). What is the average distance from an orbital electron to the nucleus for a 2p orbital in a hydrogen atom in picometers?
a) 88.2 pm
b) 317.5 pm
c) 264.6 pm
d) 105.8 pm
View Answer
Explanation: For a 2p hydrogen orbital: n = 2, l = 1, Z = 1. Plugging these into the given equation yields < r >nl = \(\frac{2^2 a_0}{Z}\)(1 + \(\frac{1}{2}\)) = 5a0 ≈ 246.6 pm
11. Expected value for radial distance from nucleus in terms of quantum numbers n and l is given by < r >nl = \(\frac{n^2 a_0}{Z} \Big\{1 + \frac{1}{2} \Big[1 – \frac{l(l+1)}{n^2}\Big]\Big\}\). What is the average distance from an orbital electron to the nucleus for a 2s orbital in a Li2+ atom in picometers?
a) 88.2 pm
b) 317.5 pm
c) 264.6 pm
d) 105.8 pm
View Answer
Explanation: For a 2s lithium ion orbital: n = 2, l = 0, Z = 3. Plugging these into the given equation yields < r >nl = \(\frac{2^2 a_0}{Z}\)(1 + \(\frac{1}{2}\)) = 2a0 ≈ 105.8 pm
12. Expected value for radial distance from nucleus in terms of quantum numbers n and l is given by < r >nl = \(\frac{n^2 a_0}{Z} \Big\{1 + \frac{1}{2} \Big[1 – \frac{l(l+1)}{n^2}\Big]\Big\}\). What is the average distance from an orbital electron to the nucleus for a 2p orbital in a Li2+ atom in picometers?
a) 88.2 pm
b) 317.5 pm
c) 264.6 pm
d) 105.8 pm
View Answer
Explanation: For a 2p lithium ion orbital: n = 2, l = 1, Z = 3. Plugging these into the given equation yields < r >nl = \(\frac{2^2 a_0}{Z}\)(1 + \(\frac{1}{2}\)(1-\(\frac{2}{4})) = \frac{5}{3}\)a0 ≈ 88.2 pm
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