chemistry : class11 Probability Density : Concept of orbitals : Structure of Atom MCQ Questions and Answers
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Probability Density
MCQ questions and answers.
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MCQ Questions
1. For a hydrogen atom in the 1s state, the wavefunction is proportional to e^(-r/a₀). What is the ratio of the probability density at r = a₀ to that at r = 0?
A.
e^(-1)
B.
e^(-2)
C.
e^(-1/2)
D.
1/e^4
Correct Answer:
e^(-2)
Explanation:
Probability density is proportional to |ψ|². Since ψ is proportional to e^(-r/a₀), the density is proportional to e^(-2r/a₀). Thus, at r = a₀ relative to r = 0, the ratio is e^(-2).
2. At a point where the wavefunction of an electron has the value ψ = -3 units, what is the corresponding relative probability density?
A.
-9 units
B.
3 units
C.
6 units
D.
9 units
Correct Answer:
9 units
Explanation:
Probability density is the square modulus of the wavefunction, |ψ|². Therefore, for ψ = -3, the probability density is (-3)² = 9 units. The negative sign of the wavefunction does not make the probability density negative.
3. Two points in an orbital have wavefunction values ψ₁ = 2 and ψ₂ = 5, with both values real. What is the ratio of the probability densities at these points, ρ₁:ρ₂?
A.
2:5
B.
4:25
C.
5:2
D.
25:4
Correct Answer:
4:25
Explanation:
Probability density is proportional to the square of the wavefunction. Hence, ρ₁:ρ₂ = ψ₁²:ψ₂² = 2²:5² = 4:25.
4. Which statement correctly distinguishes probability density from probability of finding an electron?
A.
Probability density is the probability in a finite volume, whereas probability is its value at one point.
B.
Probability density is |ψ|² at a point, while probability in a small volume is approximately |ψ|² multiplied by that volume.
C.
Probability density and probability are always numerically identical.
D.
Probability density is proportional to ψ, while probability is proportional to ψ³.
Correct Answer:
Probability density is |ψ|² at a point, while probability in a small volume is approximately |ψ|² multiplied by that volume.
Explanation:
The probability density at a point is given by |ψ|². For a sufficiently small volume element dτ, the probability of finding the electron there is |ψ|²dτ. Thus, density and probability are related but are not the same physical quantity.
5. An orbital has a nodal surface passing through a particular point. What is the probability density exactly on that nodal surface?
A.
Maximum
B.
Equal to the average density
C.
Zero
D.
Dependent only on the electron's mass
Correct Answer:
Zero
Explanation:
At a node, the wavefunction is zero. Since probability density is |ψ|², it is also zero at every point on the nodal surface.
6. For a hydrogen 2s orbital, the radial part of the wavefunction becomes zero at r = 2a₀. What is true at this spherical surface?
A.
The probability density is zero, although the orbital may have nonzero density on either side of the surface.
B.
The probability density is maximum because the radius is nonzero.
C.
The electron must be located permanently on this surface.
D.
The probability density is undefined because the wavefunction changes sign.
Correct Answer:
The probability density is zero, although the orbital may have nonzero density on either side of the surface.
Explanation:
The 2s radial wavefunction has a radial node at r = 2a₀, where ψ = 0. Therefore, the probability density |ψ|² is zero on that spherical surface, while it can be nonzero at radii smaller or larger than the node.
7. At a point in space, two wavefunctions have values ψ and -ψ, respectively. How do their probability densities compare at that point?
A.
The second density is negative relative to the first.
B.
The second density is twice the first.
C.
The two probability densities are equal.
D.
The first density is four times the second.
Correct Answer:
The two probability densities are equal.
Explanation:
Changing the sign or phase of a wavefunction does not change its modulus squared. Since |ψ|² = |-ψ|², the two wavefunctions give identical probability densities at the point.
8. For a hydrogen 2p orbital, what is the probability density exactly at the nucleus?
A.
Zero, because the 2p wavefunction vanishes at r = 0
B.
Maximum, because the nucleus is the center of the orbital
C.
Equal to the 1s probability density at the nucleus
D.
Infinite, because the electron is attracted to the nucleus
Correct Answer:
Zero, because the 2p wavefunction vanishes at r = 0
Explanation:
For a p orbital, the radial wavefunction contains a factor proportional to r near the nucleus, so ψ becomes zero at r = 0. Consequently, the probability density |ψ|² is zero at the nucleus for a 2p orbital.
9. At two locations, the probability densities are 16 units and 4 units, respectively. What is the ratio of the magnitudes of the wavefunctions at these locations?
A.
1:4
B.
2:1
C.
4:1
D.
16:4
Correct Answer:
2:1
Explanation:
Probability density is proportional to |ψ|². Therefore, the ratio of wavefunction magnitudes is the square root of the density ratio: |ψ₁|:|ψ₂| = √16:√4 = 4:2 = 2:1.
10. A wavefunction changes sign across a nodal plane but has equal magnitudes at two points located symmetrically on opposite sides of that plane. How do the probability densities at those two points compare?
A.
They are equal because probability density depends on the square modulus.
B.
The density on the negative side is negative.
C.
The density on the positive side is twice as large.
D.
Both densities are necessarily zero because they are near a node.
Correct Answer:
They are equal because probability density depends on the square modulus.
Explanation:
A sign change means the wavefunction values may be ψ and -ψ. Their probability densities are |ψ|² and |-ψ|², which are equal. The sign or phase of the wavefunction affects the wavefunction itself but not its probability density.