Engineering Physics Questions and Answers – Electron Probability Density

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This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Electron Probability Density”.

1. The electron probability density is greatest at ____________
a) r = 0
b) r = n
c) r = l
d) r = me
View Answer

Answer: a
Explanation: The probability of finding an electron is always the greatest at r = 0. As we move away from the nucleus, the probability decreases.
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2. Which of the following is the correct expression for the total number of nodes?
a) n – 1
b) l – 1
c) l + 1
d) n + 1
View Answer

Answer: a
Explanation: The principal quantum number, n, gives the shell to which the electron belongs. N – 1 gives us the total number of nodes in the orbital.

3. What orbital never has a zero probability of finding electrons?
a) s
b) px
c) dxy
d) dz2
View Answer

Answer: d
Explanation: The orbital dz2 has no nodal plane i.e., the probability of finding an electron is never zero for this orbital. For others, the nodal plane is given by n -1.

4. Nodes are the plane where the probability of finding an electron is 1.
a) True
b) False
View Answer

Answer: b
Explanation: Nodal planes are described as the planes where the probability of finding an electron is not equal to zero. The total number of nodes in an orbital is n -1.

5. If Ψ is the wave function, the probability density function is given by _____________
a) |Ψ|
b) |Ψ|2
c) |Ψ|3
d) |Ψ|4
View Answer

Answer: b
Explanation: The probability density function for a wave function, Ψ, is given by |Ψ|2. It is always greater than or equal to zero and less than or equal to one.
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6. For a 3p orbital, what are the total number of nodes?
a) 3
b) 2
c) 1
d) 0
View Answer

Answer: b
Explanation: As we know, the total number of nodes in an orbital is given by n – 1. In 3p orbital, n = 3. Thus, total number of nodes is 2.

7. The probability of finding an electron is along two directions in which orbital?
a) s
b) p
c) d
d) p
View Answer

Answer: c
Explanation: For a d-orbital, the probability of finding an electron is along two directions. For p orbitals, the probability of finding an electron is along one direction only and for an s-orbital it is uniform in every direction.

8. Arrange in the increasing number of radial nodes: 3p, 4s, 1s, 5d.
a) 1s < 3p < 4s < 5d
b) 1s < 4s < 3p < 5d
c) 1s < 3p < 5d < 4s
d) 5d < 1s < 3p < 4s
View Answer

Answer: c
Explanation: The radial nodes in an orbital is given by n – l -1. For 3p it is 2, for 5d it is 3, for 4s it is 4 and for 1s it is 1. Thus, the correct order is: 1s < 3p < 5d < 4s.

9. Which of the following can be the quantum numbers for an orbital?
a) n = 4, l = 4, m = 3
b) n = 2, l = 3, m = 1
c) n = 3, l = 2, m = -1
d) n = 3, l = 0, m = -3
View Answer

Answer: c
Explanation: In the given options, option n = 3, l = 2, m = -1 is the correct option because in this the value of l is between 0 – n-1 and value of m is between –l to +l.
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10. For which quantum number, the probability of finding an electron is most?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: The probability of finding an electron in the first orbit is maximum. As we know, s orbitals have a uniform probability of finding an electron in every direction and 1s has no nodal plane.

11. Identify the orbital for the following probability density graph.
engineering-physics-questions-answers-electron-probability-density-q11
a) 1s
b) 2s
c) 2p
d) 3s
View Answer

Answer: b
Explanation: The given figure is the probability density graph for the 2s orbital. As the graph touches the x-axis once, it means there is one nodal plane, i.e., there is one plane where the probability of finding an electron is zero.

12. S-orbitals have no angular node.
a) True
b) False
View Answer

Answer: a
Explanation: The s-orbitals have uniform probability density function in every direction. Its shape is spherical. Also, its azimuthal quantum number is zero. Therefore, all the nodes present in s-subshell are radial nodes.

13. Calculate the minimum uncertainty in the momentum of a 4He atom confined to 0.40 nm.
a) 2.02 X 10-25 kg m/s
b) 2.53 X 10-25 kg m/s
c) 2.64 X 10-25 kg m/s
d) 2.89 X 10-25 kg m/s
View Answer

Answer: c
Explanation: We know that 4He atom is somewhere in the 0.40 nm region, therefore, Δx = 0.40 nm.
Using, Δpx ≥ ℏ/Δ*x
For minimum uncertainty, Δpx = 6.626 X 10-34 Js/2π X 0.40 X 10-9
= 2.64 X 10-25 kg m/s.
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14. Which of the following is the correct expression for the magnetic moment of the electron?
a) \(\sqrt{n+1}\)
b) \(\sqrt{n(n+1)}\)
c) \(\sqrt{n(n+2)}\)
d) \(\sqrt{m(n+1)}\)
View Answer

Answer: c
Explanation: The magnetic moment of an electron is given by the expression: \(\sqrt{n(n+2)}\). Thus, when the principal quantum number, n, is zero the magnetic moment is zero as well.

15. The probability of finding an electron is zero in pxy along __________
a) x-axis
b) y-axis
c) z-axis
d) never zero
View Answer

Answer: c
Explanation: The probability of finding an electron is zero or the nodal plane is along the plane where the two lobes touch each other- which is the z-axis.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn