This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Schrodinger Equation”.

1. The Schrodinger equation is a mathematical representation for waves of which of the following fundamental balances of nature?

a) Energy Balance

b) Momentum Balance

c) Force Balance

d) Equilibrium Balance

View Answer

Explanation: The Schrodinger Equation equates energy for a wave at a certain state. The left-hand side includes all components of energy (such as potential, kinetic, and others) and the right-hand side is the total energy of a given state.

2. What is the interpretation of a wavefunction?

a) It gives the total energy of a wave

b) It helps identifying whether a particle can be treated using classical or quantum mechanics

c) The square of a wavefunction gives the probability of finding a particle at a given point in space

d) It represents the total movement of a particle over a period

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Explanation: Wavefunctions tell one about the total probability of finding a particle at positions in space. They have nothing to do with energy or time-based movement of a particle. The wavefunction has no bearing on classical or quantum treatment either.

3. Why is a wavefunction usually normalized?

a) It simplifies an otherwise complicated wave equation

b) It gives a wave’s trajectory in space

c) It gives the total energy of a wave

d) Total probability can be treated easily if normalized to one

View Answer

Explanation: Integrating the square of a wavefunction (in complex and real space) gives the total probability of finding a particle at a given point in space. Normalization of this function to one makes it easier to understand probability density and its implications. Hence this norm is used throughout quantum mechanics consistently.

4. Since the probability density is the product of the real and complex wavefunction (which is interpreted as the square of a wavefunction), can the probability be negative or non-real?

a) It can be non-real but not negative

b) It can be negative but not non-real

c) It cannot be negative or non-real

d) It is non-real in some scenarios

View Answer

Explanation: Probability density must be real and positive. It is mathematically impossible for an existing wavefunction to be imaginary or negative. For examples, if φ = a+ib & φ

^{*}= a-ib (if the wavefunction is complex), |φ|

^{2}= a

^{2}+b

^{2}, which is both real and non-negative. This property must be true for all wavefunctions

5. In the time dependent Schrodinger equation, what does V(x, t) represent?

a) Kinetic energy as a function of position and time

b) Applied force as a function of position and time

c) Applied external potential as a function of position and time

d) Voltage drop as a function of position and time

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Explanation: V(x, t) generally incorporates an external potential that could be an electric field, potential drop, gravitation field, or any other external potential that could affect time and position in the Schrodinger equation. Although voltage drop is a type of external potential, this term in the generalized equation is not always used for voltage.

6. If a time derivative of the normalization integral of a generalized wavefunction obeying the Schrodinger equation is taken, what does that imply mathematically?

a) If the derivative is zero, it obeys the normalization condition

b) If the derivative is non-zero, it obeys the normalization condition

c) If the derivative leaves a time term, it obeys the normalization condition

d) If the derivative leaves a position term, it obeys the normalization condition

View Answer

Explanation: If a function is normalized at t = 0 for an arbitrary position and satisfies the time-dependent Schrodinger equation, then the function is normalized for any later moment in time, i.e. \(\int_{-\infty}^{\infty}\) φ

^{*}(x, t)φ(x, t) = 1. Taking the time derivative on the left-hand side makes the right-hand side equal to zero, thus satisfying the normalization condition for the given criteria. If there are any time or position variables on the right-hand side, this condition is automatically violated.

7. Taking a time derivative of a normalization integral obeying Schrodinger’s equation, simplifying the normalization integral and distributing the time derivative term among the wavefunction terms gives the left-hand side as \(\frac{ih}{2m} [\frac{d\varphi(x,t)}{dx}\) φ^{*}(x,t) – \(\frac{d\varphi^*(x,t)}{dx}\) φ(x,t)\(]_{-\infty}^{\infty}\) (math is excluded for simplicity). What property of the wavefunction makes the above expression converge to zero and hence satisfy the normalization condition?

a) As x tends to infinity, the wavefunction must tend to zero as the probability of finding a particle becomes negligible at the furthest position

b) As t tends to infinity, the wavefunction must tend to zero as the probability of finding a particle becomes negligible at the farthest position

c) The imaginary term automatically converges to zero regardless of position

d) The real term automatically converges to zero regardless of position

View Answer

Explanation: An inherent wavefunction property is that φ—→ 0 as x—→ ±∞. Time tending to infinity does not matter as the derivative variables is that of position. Neither the imaginary nor the real term can converge to zero regardless of position. In that case, the wavefunction would not exist.

8. Given that the time dependent Schrodinger equation preserves the wavefunction normalization, does this mean that the time independent Schrodinger equation, i.e. φ(x), preserves this normalization criteria?

a) Yes, because the time dependent equation is a subset of the time independent equation

b) Yes, because the time dependent equation is not a subset of the time independent equation

c) No, because the time dependent equation is a subset of the time independent equation

d) No, because there is no position derivative or variables term that allows φ—→ 0 as x—→ ±∞.

View Answer

Explanation: The mathematics of normalization does not allow φ—→ 0 as x—→ ±∞ for solely a position derivative. This does not make the right-hand side of the time independent zero for any arbitrary wavefunction. Normalization criteria for the wavefunction is purely dependent on the wavefunction form itself.

9. A given wavefunction is of the form: φ = Ne\(^{-\frac{r}{a_0}}\). This is the wavefunction form of a hydrogen atom in its ground state. ‘r’ is the distance from nucleus to electron, a_{0} is the Bohr radius, and N is the normalization constant. What does this wavefunction tend to as the electron moves infinitely far from the nucleus?

a) Infinity

b) Zero

c) Negative infinity

d) One

View Answer

Explanation: \(\lim_{r \to \infty}\) Ne\(^{-\frac{r}{a_0}}\) = Ne\(^{-\frac{\infty}{a_0}}\) = N × 0 = 0 . Once an electron moves infinitely far from the nucleus, it is no longer part of the electron-nucleus system and thus the wavefunction ceases to exist.

10. A given wavefunction is of the form: φ = Ne\(^{-\frac{r}{a_0}}\). This is the wavefunction form of a hydrogen atom in its ground state. ‘r’ is the distance from nucleus to electron, a_0 is the Bohr radius, and N is the normalization constant. What is the value of N for the wavefunction to be normalized?

a) \((\frac{1}{\pi a_0^3})\)^{1/2}

b) \((\frac{\pi}{a_0^3})\)^{1/2}

c) \((\frac{1}{\pi a_0^3})\)^{2}

d) \((\frac{\pi}{a_0^3})\)^{2}

View Answer

Explanation: An atom is approximated as a sphere. The element of volume in spherical coordinates is dτ = r

^{2}sin θ dr dθ dφ. For normalization, 1 = ∫ ψψ

^{*}dτ = N

^{2}\(\int_0^\infty e^{-\frac{2r}{a_0}}r^2 dr\int_0^\pi sin \theta \int_0^{2\pi}d\phi\). From integral tables, or by using integration by parts, \(\int_0^\infty r^2 e^{-\frac{2r}{a_0}}dr = \frac{a_0^3}{3}\). This gives 1 = N

^{2}\(\frac{a_0^3}{2}\) 2π-→ N = \((\frac{1}{\pi a_0^3})\)

^{1/2}

11. What is the mathematical meaning of orthogonality of a wavefunction?

a) ∫ φφ^{*} = 1

b) ∫ φ/φ^{*} = 1

c) ∫ φφ^{*} = 0

d) ∫ φ/φ^{*} = 0

View Answer

Explanation: A wavefunction is orthogonal if the probability of finding a particle at any point in space is zero. Physically speaking, orthogonality occurs when two or more parts of a wave cancel each other, resulting in no wave, hence non-existent probability density.

12. For a given wavefunction φ(x), what is the probability of finding a particle between points x_{1} and x_{2} on the positive x axis?

a) \(\int_0^\infty\) φ^{*}(x) φ(x)dx

b) \(\int_{-\infty}^\infty\) φ^{*}(x) φ(x)dx

c) \(\int_{x_1}^{x_2}\) φ(x)dx

d) \(\int_{x_1}^{x_2}\) φ^{*}(x) φ(x)dx

View Answer

Explanation: For the probability of finding a particle within two specified points, the probability density needs to be integrated between those two specific points, without any infinity. Here, the two points are x

_{1}and x

_{2}.

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