This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Eigenfunctions and Probability Densities for Hydrogen Like Atoms”.

1. What is the wavefunction for a one-electron system called?

a) An eigenvalue

b) A vector

c) An eigenfunction

d) An orbital

View Answer

Explanation: Orbitals are quantum mechanical wavefunctions for single electron systems. There are three types of atomic orbitals: s, p, d, and f. Each of these have a shape and type of wavefunction describing them. Orbitals are associated with their quantum numbers n, l, and m.

2. What do radial functions for hydrogen like atoms depend on?

a) Quantum numbers n, l, and m

b) Quantum numbers n, l and atomic number Z

c) Quantum numbers n and l only

d) Atomic number Z only

View Answer

Explanation: Radial functions specifically depend on principal energy levels and angular momentum quantum numbers that determine orbital shape. Atomic number affects atomic size, and hence the radial functions that describe these atomic shapes.

3. A hydrogen atom described by the wavefunction: ψ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is the expectation value of the total angular momentum operator, L^{2}?

a) 5h^{2}

b) h^{2}

c) 3h^{2}

d) 2h^{2}

View Answer

Explanation: L

^{2}is given by the formula: L

^{2}= ∑|C

_{n}|

^{2}h

^{2}l(l+1), where l is the angular quantum number and C

_{n}is the coefficient of each superposition state. For the given superposition, l is one and 2 respectively, which gives L

^{2}= \((\frac{1}{2})^2\) h

^{2}(1(1 + 1)) + \((\frac{\sqrt 3}{2})^2\) h

^{2}(2(2 + 1)) = 5h

^{2}

4. A hydrogen atom described by the wavefunction: ψ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} > What is the expectation value for the energy in this state?

a) –\(\frac{1}{10}\)R_{H}

b) –\(\frac{47}{230}\)R_{H}

c) –\(\frac{43}{576}\)R_{H}

d) –\(\frac{2}{5}\)R_{H}

View Answer

Explanation: Total expectation of energy is given by ∑|C

_{n}|

^{2}\((\frac{-R_H}{n^2})\). This gives the total energy as \((\frac{1}{4} × \frac{-R_H}{3^2}) + (\frac{3}{4} × \frac{-R_H}{4^2}) = -\frac{1}{36}\) R

_{H}–\(\frac{3}{96}\) R

_{H}= –\(\frac{43}{576}\) R

_{H}.

5. A hydrogen atom described by the wavefunction: ψ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is the expectation value of L_{z}, the projection of angular momentum on the z-axis?

a) -1.75h

b) 1.75h

c) 2h

d) -2h

View Answer

Explanation: Total expectation of L

_{z}is given by ∑|C

_{n}|

^{2}(mh), where ‘m’ denotes the magnetic quantum number of the state. For the 3p

_{z}orbital, m = -1 and for the 4d

_{-2}orbital m = -2. This gives L

_{z}= \((\frac{1}{4}\) × -1 × h) + \((\frac{3}{4}\) × -2 × h) = -1.75h

6. A hydrogen atom has one radial plane and two radial nodes. What are the values of n and l?

a) n = 2, l = 1

b) n = 2, l = 0

c) n = 3, l = 0

d) n = 4, l = 1

View Answer

Explanation: A 4p orbital has one radial plane and two radial nodes. The number of radial nodes is equal to n-l-1. Since only a p orbital can have one radial plane, l must be 1, and plugging this in the previous expression gives n = 4 and l = 1.

7. A hydrogen atom has one radial plane and two radial nodes. How many orbitals fit this description?

a) 4 orbitals

b) 2 orbitals

c) 1 orbital

d) 3 orbitals

View Answer

Explanation: The relationship between quantum number(s) and number of radial nodes is given by #radial nodes = n-l-1. L is 1 since a “p” orbital can only have a single radial plane and plugging it into the expression gives n as 4. For a principal quantum number of 4 and a p orbital, there are 3 possible orbitals of 4p

_{x}, 4p

_{y}, or 4p

_{z}. Hence the description fits 3 orbitals.

Since the orbital is 4p, it can be 4p

_{x}, 4p

_{y}, or 4p

_{z}. Thus, this description fits 3 orbitals.

8. A hydrogen atom exists in a state |4d_{xy} >. Is this an eigenstate of the Hamiltonian, and if yes, what is its eigenvalue?

a) Yes, eigenvalue: –\(\frac{R_H}{16}\)

b) Yes, eigenvalue: –\(\frac{R_H}{9}\)

c) Yes, eigenvalue: –\(\frac{R_H}{4}\)

d) No, it is not an eigenstate of the Hamiltonian

View Answer

Explanation: Energy levels are given by E

_{n}= –\(\frac{R_H}{n^2}\). The principal quantum number for this state is n = 4, hence the eigenvalue for this state is E

_{n}= –\(\frac{R_H}{n^2}\) = –\(\frac{R_H}{16}\)

9. A hydrogen atom exists in a state |4d_{xy} >. Is this an eigenstate of the angular momentum operator L^{2}, and if yes, what is its eigenvalue?

a) Yes, eigenvalue: 2ħ^{2}

b) Yes, eigenvalue: 6ħ^{2}

c) Yes, eigenvalue: 3πħ^{2}

d) No, it is not an eigenstate of this operator

View Answer

Explanation: Since l=2 in this state (d orbital), Eigenstates for the angular momentum operator for this is given by L

^{2}= ħ

^{2}l(l + 1) = 6ħ

^{2}

10. A hydrogen atom exists in a state |4d_{xy} >. Is this an eigenstate of the angular momentum operator projection L_{z}, and if yes, what is its eigenvalue?

a) Yes, eigenvalue: 2ħ

b) Yes, eigenvalue: 6ħ

c) Yes, eigenvalue: 3ħ

d) No, it is not an eigenstate of this operator

View Answer

Explanation: This is not an eigenstate because this hydrogen atom is a superposition of states with different quantum numbers m: |4d

_{xy}> = \(\frac{1}{\sqrt 2i}\)(|4d

_{+2}> – |4d

_{-2}>)

11. A hydrogen atom exists in the superposition state: φ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is one possible outcome of measuring L_{z} with its probability?

a) 2ℏ, probability: 0.5

b) ℏ, probability: 0.75

c) 0, probability: 0.5

d) 0, probability: 0.25

View Answer

Explanation: 3p

_{z}is equivalent to 3p

_{0}with n = 3, l = 1 and m = 0. For m = 0, L

_{z}= mℏ = (0)ℏ = 0. Probability = (\(\frac{1}{2})^2 = \frac{1}{4}\).

12. A hydrogen atom exists in the superposition state: φ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is one possible outcome of measuring L_{z} with its probability?

a) -2ℏ, probability: 0.75

b) ℏ, probability: 0.75

c) 0, probability: 0.5

d) 0, probability: 0.75

View Answer

Explanation: 4d

_{-2}is n = 4, l = 2 and m = -2. For m = -2, L

_{z}= mℏ =(-2)ℏ = -2ℏ. Probability = \((\frac{\sqrt 3}{2})^2\) = 0.75

13. A hydrogen atom exists in the superposition state: φ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is the expectation value of L_{z} in this state?

a) ℏ

b) -1.5ℏ

c) -2ℏ

d) 0

View Answer

Explanation: < L

_{z}> = P

_{1}.0 + P

_{2}.(-2ℏ) = ℏ(0 – 1.5) = -1.5ℏ.

14. A hydrogen atom exists in the superposition state: φ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is the expectation value of L^{2} in this state?

a) ℏ^{2}

b) -1.5ℏ

c) 5ℏ^{2}

d) 0

View Answer

Explanation: < L

^{2}> = 0.25ℏ

^{2}1(1 + 1) + 0.75ℏ

^{2}2(2 + 1) = ℏ

^{2}(0.5 + 4.5) = 5ℏ

^{2}.

15. A hydrogen atom exists in the superposition state: φ = \(\frac{1}{2}\)|3p_{z} > + \(\frac{\sqrt 3}{2}\)| 4d_{-2} >. What is the expectation value of energy in this state?

a) \(\frac{97}{433}\)R_{H}

b) –\(\frac{43}{576}\)R_{H}

c) 0

d) \(\frac{R_H}{2}\)

View Answer

Explanation: < E > = P

_{1}.(\(\frac{-R_H}{9})\) + P

_{2}.(-\(\frac{R_H}{16})\) = -R

_{H}(\(\frac{1}{36} + \frac{3}{64}) = -\frac{43}{576}\)R

_{H}.

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