This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Variational Method”.
1. The variational method is given by ∫ φ* Hφ dτ ≥ Egs. What Egs signify?
a) True ground state energy
b) Energy level lower than the state of given wavefunction
c) Highest possible energy state for given wavefunction
d) Hamiltonian operator
View Answer
Explanation: Egs is the true ground state energy. The use of an approximate wavefunction always yields an energy value higher than the ground state energy. The wavefunction φ is iterated till the lowest possible error is obtained between the lest hand side of this equation and the ground state energy.
2. How are the values for integration constants obtained when solving for set of functions E = \(\frac{\int \varphi^* H\varphi d\tau}{\int \varphi^* \varphi d\tau}\)?
a) By comparing values with Egs
b) Using boundary conditions for the given system
c) Using wavefunction constraints
d) By solving the simultaneous equations \(\frac{\partial E}{\partial C_i}\) = 0
View Answer
Explanation: Solving the given equation will yield a set of constants that can only be found if the energy derivative with respect to each constant is taken and solved simultaneously. This will give energy eigenvalues for the energy state.
3. The variational method is used to find energy levels for a particle in a box. The wavefunction is given by φ = \(\frac{30}{a^{\frac{5}{2}}}\)x(a – x). What are the boundary conditions for this system?
a) φ = 0 at x = 0, x = a
b) φ = ∞ at x = 0
c) φ = 0 at x = a
d) φ = 0 at x = \(\frac{a}{2}\)
View Answer
Explanation: Like a normal particle in a box, the boundary conditions remain the same. The wavefunction as to be 0 at the boundary edges for it to be continuous. This is because the wavefunction is 0 for x<0 and x>a. The potential energy operator, as part of the Hamiltonian, is infinite in these regimes.
4. A trial wavefunction for a particle in a box is given by φ = \(\frac{30}{a^{\frac{5}{2}}}\)x(a – x). What is the ground state energy of this using the variational method?
a) \(\frac{4\pi \varepsilon_0}{h^2}\)
b) 0
c) \(\frac{h^2}{8ma^2}\)
d) \(\frac{5h^2}{4\pi^2 ma^2}\)
View Answer
Explanation: \(\int_0^a\) φ*Hφdτ = \(\frac{-30\hbar^2}{2a^5 m} \int_0^a\)(ax-x2)\(\frac{d^2}{dx^2}\)(ax – x2)dx = \(\frac{-30\hbar^2}{a^5 m} \int_0^a\)(x2 – ax)dx = \(\frac{5h^2}{4\pi^2 ma^2}\).\(\frac{h^2}{8ma^2}\) is the true ground state energy when the Schrodinger equation for a particle in the box is solved.
5. A trial wavefunction for a particle in a box is given by φ = \(\frac{30}{a^{\frac{5}{2}}}\)x(a – x). What is the percent error in ground state energy using this variational method VS the actual Schrodinger equation solution?
a) 84.0 %
b) 49.1 %
c) 1.3%
d) 14.8%
View Answer
Explanation: \(\int_0^a\) φ*Hφdτ = \(\frac{-30\hbar^2}{2a^5 m} \int_0^a\)(ax-x2)\(\frac{d^2}{dx^2}\)(ax – x2)dx = \(\frac{-30\hbar^2}{a^5 m} \int_0^a\)(x2 – ax)dx = \(\frac{5h^2}{4\pi^2 ma^2}\).\(\frac{h^2}{8ma^2}\) is the true ground state energy when the Schrodinger equation for a particle in the box is solved. Hence, % error = \(\frac{(\frac{5}{4} \pi^2)-\frac{1}{8}}{\frac{1}{8}}\) × 100 = 1.3%. There is no variation calculation as there is no variation parameter in this case.
6. A particle of mass m is moving in a 1D potential V(x) = b|x|. Ground state wavefunction to vary is given by φ = \((\frac{\alpha}{\pi})^{1/4}\) e\(^{-\frac{\alpha x^2}{2}}\). What is the value of the variational integral W(∝) = < φ|H|φ >?
a) \(\frac{\alpha \hbar^2}{4m} + \frac{b}{\sqrt {\alpha \pi}}\)
b) \(\frac{b}{\sqrt {\alpha \pi}}\)
c) \(\frac{\alpha b}{4m}\)
d) 0
View Answer
Explanation: W = < φ0|T|φ0 > + < φ0|V|φ0 > = –\(\frac{\hbar m\omega}{2.2m}\) < φ0|(a – a∔)2|< φ0 > + b\((\frac{\alpha}{\pi})^{\frac{1}{2}}\) 2\(\int_0^\infty\) xe-αx2 = \(\frac{-\hbar \omega}{4}\)(-1) + \(\frac{b}{\alpha}(\frac{\alpha}{\pi})^{\frac{1}{2}}\) = \(\frac{\alpha \hbar^2}{4m} + \frac{b}{\sqrt {\alpha \pi}}\).
7. A particle of mass m is moving in a 1D potential V(x) = b|x|. Ground state wavefunction to vary is given by φ = \((\frac{\alpha}{\pi})^{1/4}\) e\(^{-\frac{\alpha x^2}{2}}\). What is the minimum value of the α, found by minimizing the variation integral W, where W(∝) = < φ|H|φ >?
a) αmin = (\(\frac{bm}{\hbar^2 \sqrt \pi})^{\frac{2}{3}}\)
b) αmin = ℏ2√πbm
c) αmin = 0
d) αmin = \(\frac{3}{2}\)α2ℏπ
View Answer
Explanation: W = < φ0|T|φ0 > + < φ0|V|φ0 > = –\(\frac{\hbar m\omega}{2.2m}\) < φ0|(a – a∔)2| φ0 > + b(\(\frac{\alpha}{\pi})^{\frac{1}{2}}\) 2\(\int_0^\infty\) xe-αx2 = \(\frac{-\hbar \omega}{4}\)(-1) + \(\frac{b}{\alpha}(\frac{\alpha}{\pi})^{\frac{1}{2}}\) = \(\frac{\alpha \hbar^2}{4m} + \frac{b}{\sqrt {\alpha \pi}}\) . For min, \(\frac{dW}{d\alpha} = 0 \rightarrow \frac{dW}{d\alpha} = \frac{\hbar^2}{4m} – \frac{b}{2\sqrt{\alpha^3 \pi}}\). This gives αmin = (\(\frac{bm}{\hbar^2 \sqrt \pi})^{\frac{2}{3}}\).
8. A particle of mass m is moving in a 1D potential V(x) = b|x|. Ground state wavefunction to vary is given by φ = (\(\frac{\alpha}{\pi})^{1/4}\) e\(^{-\frac{\alpha x^2}{2}}\). What is the best approximation to the ground state energy E0?
a) 9πℏ2
b) \(\frac{2\pi}{bm}\)
c) \(\frac{5}{4}(\frac{b^2 \hbar^2}{\pi m})^{\frac{1}{3}}\)
d) \(\frac{9}{8}\frac{bm}{\pi}\)
View Answer
Explanation: W = < φ0|T|φ0 > + < φ0|V|φ0 > = –\(\frac{\hbar m\omega}{2.2m}\) < φ0|(a – a∔)2| φ0 > + b\((\frac{\alpha}{\pi})^{\frac{1}{2}}\) 2\(\int_0^\infty\) xe-αx2 = \(\frac{-\hbar \omega}{4}\)(-1) + \(\frac{b}{\alpha}(\frac{\alpha}{\pi})^{\frac{1}{2}}\) = \(\frac{\alpha \hbar^2}{4m} + \frac{b}{\sqrt \alpha \pi}\). For min, \(\frac{dW}{d\alpha} = 0 \rightarrow \frac{dW}{d\alpha} = \frac{\hbar^2}{4m} – \frac{b}{2\sqrt{\alpha^3 \pi}}\). This gives αmin = (\(\frac{bm}{\hbar^2 \sqrt \pi})^{\frac{2}{3}}\).E0 = W(αmin) = (\(\frac{bm}{\hbar^2 \sqrt \pi})^{\frac{2}{3}} \frac{\hbar^2}{4m} + \frac{b}{\sqrt\pi}(\frac{\hbar^2 \sqrt \pi}{bm})^{1/3} = \frac{5}{4}(\frac{b^2 \hbar^2}{\pi m})^{\frac{1}{3}}\).
9. What is the total Hamiltonian for a particle in motion with initial Hamiltonian H0, and then experiencing a magnetic field of strength B?
a) H0 + \(\frac{eB}{2m_e}\)Lz
b) H0
c) 0
d) \(\frac{eB}{2m_e}\)Lz
View Answer
Explanation: The total Hamiltonian is the sum of the initial Hamiltonian and energy due to the magnetic field. This energy is the spin magnetic moment in a field of strength B. This considers the magnetic quantum number, m, and field strength. Eigenvalues for this can be found as well.
10. What are the energy eigenvalues for a particle in motion with initial Hamiltonian H0, and then experiencing a magnetic field of strength B?
a) En,l,m = \(\frac{eB\hbar}{2m_e}\)(ml + gems)
b) En,l,m = \(\frac{Bm_e}{4\pi}\)
c) En,l,m = \(\frac{m_e e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\)
d) En,l,m = \(\frac{m_e e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\) + \(\frac{eB\hbar}{2m_e}\)(ml + gems)
View Answer
Explanation: The Hamiltonian incorporates all types of energies that act upon a quantum mechanical system. This system includes magnetic energy, given by H = H0 + \(\frac{eB}{2m_e}\)Lz. Plugging this into Schrodinger’s equation and solving for eigenvalue E gives En,l,m = \(\frac{m_e e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\) + \(\frac{eB\hbar}{2m_e}\)(ml + gems) This is given by H = H0 + \(\frac{eB}{2m_e}\)Lz. Plugging this into Schrodinger’s equation and solving for eigenvalue E gives En,l,m = \(\frac{m_e e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\) + \(\frac{eB\hbar}{2m_e}\)(ml + gems).
11. How is the total angular momentum of a system J described in terms of spin angular momentum S and orbital angular momentum L?
a) J = L
b) J = L+S
c) J = 2πL + SBe\(^{-\frac{\hbar^2}{2}}\)
d) J = S2L
View Answer
Explanation: Total angular momentum is just a vector sum of spin angular momentum and orbital angular momentum. This leads to a phenomenon called spin-orbit coupling that contributes to the total Hamiltonian of a system. In this case, total angular momentum of the atomic system is described in this way.
12. If a trial wavefunction has been changed once, and the new variational integral is smaller than the previous one, what should one do to proceed?
a) Keep changing the wavefunction in the same way
b) Change the wavefunction in a different way
c) Keep iterating both the wavefunction, as well as the variation integral until the calculated ground state energy matches the true ground state energy
d) Choose a different trial wavefunction
View Answer
Explanation: If the variation integral is getting smaller, it means the ground state energy is converging to its true value. Continuously performing these steps will reduce error in the calculated ground state energy. Putting this in a loop on MATLAB could automatically solve this.
13. If a trial wavefunction has been changed once, and the new variational integral is larger than the previous one, what should one do to proceed?
a) Keep changing the wavefunction in the same way
b) Change the wavefunction in a different way
c) Keep iterating both the wavefunction, as well as the variation integral until the calculated ground state energy matches the true ground state energy
d) Choose a different trial wavefunction
View Answer
Explanation: If the variation integral is getting smaller, it means the ground state energy is diverging from its true value. This means that the method of iteration is incorrect. Trying different ways to change this will help in finding a better convergence condition.
14. Can the variational method be used to calculate energy states when there has been a disturbance to the system?
a) Yes, because the new Hamiltonian and wavefunctions should account for these disturbances
b) No, the disturbance will not change the energy levels
c) No, because new iterated functions rely on previous functions that do not include these disturbances
d) Yes, because the disturbances will not change the energy levels
View Answer
Explanation: Disturbances can change energy levels significantly and this must be considered. A new method called Perturbation theory is used to account for these disturbances. It is a modification of the variational principle.
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