This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Helium Atom”.
1. How is the Helium atom described?
a) A centered nucleus surrounded by two electrons
b) A centered electron surrounded by a nucleus and another electron
c) A centered proton surrounded by two electrons
d) A centered neutron surrounded by two electrons
View Answer
Explanation: A helium atom consists of a nucleus (which is taken as the origin coordinate for convenience) surrounded by two electrons. The nucleus in term has one proton and one neutron. Distances between each particle are recorded with algebraic expressions.
2. Assuming the nucleus is fixed at the origin, distances between each electron and the nucleus is r1 & r2 respectively, and distance between both electrons is r12, how is the Hamiltonian for this system written?
a) H = \(\frac{-\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2 + \nabla_{nucleus}^2) – \frac{1}{4\pi \varepsilon_0}(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} -\frac{e^2}{r_{12}})\)
b) H = –\(\frac{1}{4\pi \varepsilon_0}(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} -\frac{e^2}{r_{12}})\)
c) H = 0
d) H = \(\frac{-\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) – \frac{1}{4\pi \varepsilon_0}(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} -\frac{e^2}{r_{12}})\)
View Answer
Explanation: The Hamiltonian in this case is the sum of kinetic and potential energies of each electron and the nucleus. The Laplacian square of the nucleus is ignored because it is fixed at the origin, hence not having a kinetic energy. The potential energy of each electron with each other must be considered as well, hence the \(\frac{e^2}{r_{12}}\) term.
3. If the electrons do not interact with each other, what is the Hamiltonian of this Helium atom system?
a) H = \(\frac{-\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2 + \nabla_{nucleus}^2) – \frac{1}{4\pi \varepsilon_0}(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} -\frac{e^2}{r_{12}})\)
b) H = –\(\frac{1}{4\pi \varepsilon_0}(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} -\frac{e^2}{r_{12}})\)
c) H = 0
d) H = \(\frac{-\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) – \frac{1}{4\pi \varepsilon_0}(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} -\frac{e^2}{r_{12}})\)
View Answer
Explanation: Since the electrons do not interact with each other, the \(\frac{e^2}{4\pi \varepsilon_0 r_{12}}\) term is neglected. This manifests itself as the potential energy of interaction. This is the first approximation in solving the Schrodinger equation for this system to get an exact solution.
5. If the electrons do not interact with each other, how can the wavefunction for the helium atom be written in terms of orbital electrons?
a) φ = 2s(1)1s(2)
b) φ = 1s(2)2s(1)
c) φ = 1s(1)1s(2)
d) φ = 1s(1)2s(1)
View Answer
Explanation: The wavefunction is written as a product of the orbital wavefunctions of the 1s(1) and 1s(2) electrons. This is an approximate wavefunction and can be expanded to φ = \(\frac{1}{\pi}(\frac{Z}{a_0})^3 e^{\frac{-Zr_1}{a_0}} e^{-\frac{Zr_2}{a_0}}\). This is the ground state wavefunction and can be used as a trial function for the variational method.
6. The effective energy of a helium atom is given as E’ = [(Z’)2 – \(\frac{27Z^{‘}}{8}] \frac{e^2}{4\pi \varepsilon_0 a_0}\). What is the minimum value of effective nuclear charge Z’?
a) 1
b) 27/16
c) 47/88
d) 2
View Answer
Explanation: Differentiating the above expression gives \(\frac{dE^{‘}}{dZ^{‘}}\) = (2Z’ – \(\frac{27}{8})\frac{e^2}{4\pi \varepsilon_0 a_0}\). Equating this to zero and re-arranging gives Z’ = Zmin = \(\frac{27}{16}\). This implies the effective nuclear charge of the helium atom is lesser than 2 because of the shielding effect of electrons.
7. The effective energy of a helium atom is given as E’ = [(Z’)2 – \(\frac{27Z^{‘}}{8}\)] \(\frac{e^2}{4\pi \varepsilon_0 a_0}\). What is the minimum value of this energy E’?
a) -77.5 eV
b) 77.5 eV
c) 108.23 eV
d) -108.23 eV
View Answer
Explanation: Differentiating the above expression gives \(\frac{dE^{‘}}{dZ^{‘}}\) = (2Z’ – \(\frac{27}{8})\frac{e^2}{4\pi \varepsilon_0 a_0}\). Equating this to zero and re-arranging gives Z’ = Zmin = \(\frac{27}{16}\). Back-plugging this into the energy expression gives \(E_{min}^{‘} = \Big[(\frac{27}{16})^2 – \frac{27*\frac{27}{16}}{8}\Big] \frac{e^2}{4\pi \varepsilon_0 a_0}\) = -77.5eV.
8. If the exact energy of the ground state helium atom is -79.0 eV, what is the percent error in energy obtained by using the variational method? The effective energy is given by E’ = [(Z’)2 – \(\frac{27Z^{‘}}{8}] \frac{e^2}{4\pi \varepsilon_0 a_0}\).
a) 43%
b) 19%
c) 10%
d) 1.9%
View Answer
Explanation: Differentiating the above expression gives \(\frac{dE^{‘}}{dZ^{‘}}\) = (2Z’ – \(\frac{27}{8})\frac{e^2}{4\pi \varepsilon_0 a_0}\). Equating this to zero and re-arranging gives Z’ = Zmin = \(\frac{27}{16}\). Back-plugging this into the energy expression gives \(E_{min}^{‘} = \Big[(\frac{27}{16})^2 – \frac{27*\frac{27}{16}}{8}\Big] \frac{e^2}{4\pi \varepsilon_0 r_12}\) = -77.5eV. % error = \(\frac{79 – 77.5}{77.5}\) = 1.9%.
9. In the variational method, what parameter should be varied to solve for energy of a helium atom?
a) Effective nuclear charge
b) Effective energy
c) Bohr’s radius
d) Permittivity of free space
View Answer
Explanation: The parameters to be varied is effective nuclear charge Z’, since this needs to be plugged into the energy equation to solve for minimum energy. Moreover, this is the only variable other than E’. If E’ is used as the variable parameter, there will be no equation to plug this into to solve for minimum energy.
Sanfoundry Global Education & Learning Series – Physical Chemistry.
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