Quantum Chemistry Questions and Answers

Answer: D

The slope will be the same for all metals and independent of intensity of radiation.

Einstein's equation is

\[ KE_{\max} = h\nu - \phi \]

A plot of KE against ν is a straight line of slope h, Planck's constant, for every metal. The metal changes only the work function φ, which moves the intercept and the threshold frequency, not the slope. Intensity changes only the number of photoelectrons.

That rules out options A, B and C, which all make the slope depend on the metal or the intensity.

Answer: D

For a simple harmonic oscillator, the number of nodes of the wave function increases with increase in energy.

The wave function of level v has exactly v nodes: none for v = 0, one for v = 1, and so on, the same pattern as for a particle in a box.

The levels E_v = (v + 1/2)hν are equally spaced, so the spacing neither increases (B) nor decreases (C); decreasing spacing belongs to the anharmonic oscillator. The potential energy V = kx²/2 rises with the square of the displacement, not linearly (A).

Answer: A

The correct statement is a possible trial wave function for the ion is ψ = C₁1S_A ± C₂1S_B.

In the LCAO treatment of H₂⁺, the single electron's wave function is a linear combination of the 1s orbitals on nuclei A and B. The plus combination gives the bonding orbital and the minus combination the antibonding one.

Because the two nuclei are identical, symmetry makes the coefficients equal in magnitude (C₁ = ±C₂), so (B) is wrong. A ratio of orbitals (C) is not a linear combination, and the 1s orbitals used are normalised hydrogen functions (D).

Answer: A

The ratio is (m_p/m_e)^1/2.

A particle of charge q accelerated through a potential V gains kinetic energy qV. The electron and the proton carry the same magnitude of charge, so they gain the same kinetic energy. Its momentum and de Broglie wavelength are

\[ \begin{aligned} p &= \sqrt{2mqV} \\ \lambda &= \frac{h}{p} = \frac{h}{\sqrt{2mqV}} \end{aligned} \]

With h, q and V the same for both, λ is proportional to 1/√m:

\[ \begin{aligned} \frac{\lambda_{e}}{\lambda_{p}} &= \sqrt{\frac{m_{p}}{m_{e}}} \\ &\approx \sqrt{1836} \approx 43 \end{aligned} \]

m_e/m_p is the inverse of the mass ratio and gives the wrong direction (the lighter electron has the longer wavelength). m_p/m_e without the square root holds for equal speeds, not equal energies, and (m_p/m_e)² has no basis.

Answer: C

The degeneracies are 1, 3 and 6.

For a cubic box the bracket stands for the sum of squares of the three quantum numbers, n_x² + n_y² + n_z² (the paper prints the subscripts x, y, z beneath each n²). Every arrangement of the same three numbers gives the same energy, so the degeneracy is the number of distinct permutations:

• (1, 1, 1): sum 3; only one arrangement, so degeneracy 1.
• (2, 1, 1): sum 6; the 2 can be in any of three places, (2,1,1), (1,2,1), (1,1,2), so degeneracy 3.
• (3, 2, 1): sum 14; three different numbers can be arranged in 3! = 6 ways, so degeneracy 6.

No other set of positive integers has squares summing to 14 or 6, so there is no accidental degeneracy to add. 3, 4, 6 and 1, 3, 3 miscount the arrangements, and 1, 1, 1 treats every level as non-degenerate, which is true only for a box with three unequal sides.

Answer: C

The correct expression is (C), the product of the radial integral and the two angular integrals.

The hydrogen 1s function is \(\psi = (\pi a_{0}^{3})^{-1/2} e^{-r/a_{0}}\). The average distance is \(\langle r \rangle = \int \psi^{*} r \psi\, d\tau\), with the volume element \(d\tau = r^{2}\sin\theta\, dr\, d\theta\, d\phi\). The extra r² turns r into r³, and \(\psi^{*}\psi\) gives \(e^{-2r/a_{0}}\), so the integral separates into a radial part times \(\int_{0}^{\pi}\sin\theta\, d\theta = 2\) times \(\int_{0}^{2\pi} d\phi = 2\pi\). That is (C).

\[ \begin{aligned} \langle r \rangle &= \frac{4\pi}{\pi a_{0}^{3}} \cdot \frac{3!}{(2/a_{0})^{4}} \\ &= \frac{3}{2}a_{0} \end{aligned} \]

(A) and (B) leave out the angular integrals (and (A) has a stray minus sign), so they are short by a factor of 4π. (D), 5/2 a₀, is not the 1s value, which is 3/2 a₀.

Answer: D

It could be the equation for two charged particles in 1-D box problem.

Read the equation term by term. The kinetic energy term carries (1/m_e + 1/m_h), which is 1/μ, the reciprocal of the reduced mass of two particles, so two particles are moving relative to each other. It has only d²/dx², so the motion is in one dimension. The potential term varies as z²/r, a Coulomb interaction between charges. Together these describe two charged particles, such as an electron and a hole, confined in one dimension.

A single particle in a box has one mass and no potential inside the box. A 2D box would need derivatives in two coordinates. Simple harmonic motion would need a potential proportional to x², not to 1/r.

Answer: B

The keyed Slater determinant is (B).

For helium's ground state, both electrons occupy 1s, one with spin α and one with β. The antisymmetric wavefunction is a 2 × 2 determinant built from the spin-orbitals 1sα and 1sβ, with electron 1 and electron 2 as the other index, and the normalising factor 1/√2!. Expanding (B):

\[ \begin{aligned} \Psi &= \tfrac{1}{\sqrt{2}}\,1s(1)1s(2) \\ &\times [\alpha(1)\beta(2) - \beta(1)\alpha(2)] \end{aligned} \]

a symmetric space part times the antisymmetric singlet spin function. Writing electrons by columns rather than rows does not change a determinant.

(A) swaps the two columns and (C) puts a minus sign in front; each equals −Ψ, which differs only by an overall sign. The key takes (B), the conventional form with the positive normalisation constant and electron 1 first. (D) is wrong in substance: its 1/2 is not the right normalisation, and its entries mix the electron labels inside the spin functions.

Answer: B

The energy is an upper bound to E_o.

The variation theorem states that for any well-behaved trial function φ, the expectation value of the Hamiltonian is never lower than the true ground-state energy:

\[ \begin{aligned} E &= \frac{\int \phi^{*}\hat{H}\phi\, d\tau}{\int \phi^{*}\phi\, d\tau} \\ &\ge E_{o} \end{aligned} \]

Equality holds only if φ is the exact ground-state function. That is why the variation method improves a trial function by minimising E: the lower the energy, the closer it is to the truth.

It is not always equal to E_o (that needs the exact function), it is never a lower bound, and it need not be positive; bound-state energies such as hydrogen's are negative.

Answer: A

Born's interpretation is that the probability of finding the particle between r and r + dr is proportional to |ψ|² dr.

Max Born (1926) proposed that the wavefunction itself has no direct physical meaning, but the square of its modulus, ψ*ψ, is a probability density. That is why acceptable wavefunctions must be finite, single-valued and normalisable: the total probability over all space must equal 1.

Continuity and normalisation are conditions placed on ψ because of the Born interpretation, not the interpretation itself. The mathematical representation of an observable is an operator, which is a different postulate of quantum mechanics.