Question 1

Hydrogen atom: What is the radius of the smallest electron orbit in a singly-ionized helium atom?

Accepted Answer

The radius of an electron orbit in a hydrogen atom is given by the Bohr radius, which is equal to 0.529 Å. In a singly-ionized helium atom (He+), there is only one electron remaining. The radius of the smallest electron orbit in He+ can be calculated using the formula r = (n^2 / Z) * a0, where n is the principal quantum number (n=1 for the smallest orbit), Z is the nuclear charge (Z=2 for He+), and a0 is the Bohr radius. Plugging in these values gives r = (1^2 / 2) * 0.529 Å, which simplifies to r = 0.265 Å, or 2.65 × 10^-11 m (choice B).

Question 2

Multielectron atoms: The only VALID electron state and shell designation among the following is

Accepted Answer

The valid electron state and shell designation is 2p, L. The 2p state corresponds to the second energy level (n=2) and the p subshell, which can exist in the L shell. Other options either have incorrect principal quantum numbers or subshells that do not exist for the given shell.

Question 3

Multielectron atoms: An atom has completely filled inner shells and a single valence electron in an excited p state. The filled inner shells have an orbital momentum equal to zero. A magnetic field is applied, defining the z-axis along the field. Which of the following sets of angles are possible angles between the magnetic field and the orbital angular momentum?

Accepted Answer

The orbital angular momentum of an electron in a p state can take on orientations that correspond to the magnetic quantum numbers $m_l = -1, 0, +1$. These orientations relate to different angles with respect to the magnetic field direction, including 45°, 90°, and 135°, making all these angles possible.

Question 4

Multielectron atoms: Consider the n = 9 shell.
(a) What is the largest value of the orbital quantum number, l, in this shell?
(b) How many electrons can be placed in this shell?

Accepted Answer

The answer of Multielectron atoms: Consider the n = 9...

Question 5

Multielectron atoms: An atom with 5 electrons is in its ground state. How many electrons are in its outermost shell?

Accepted Answer

The answer of Multielectron atoms: An atom with 5 electrons...

Question 6

Multielectron atoms: What is the electron configuration for ground state Li, which has 3 electrons?

Accepted Answer

The electron configuration for ground state Li would be 1s²2s¹. Using spectroscopic notation, this would be written as 1s²2s¹. Choice C is the only option that includes this electron configuration.

Question 7

Multielectron atoms: How many electrons can be found with principal quantum number  in a suitably heavy atom?

Accepted Answer

The answer of Multielectron atoms: How many electrons can be...

Question 8

Angular momentum: The magnitude of the orbital angular momentum L of an electron in a certain atom is equal to 3.464ħ. What is the orbital quantum number l of the electron?

Accepted Answer

The magnitude of the orbital angular momentum L is given by the formula L = √(l(l+1))ħ. Solving for l when L = 3.464ħ gives l = 3.

Question 9

Angular momentum: The magnitude of the orbital angular momentum L of an electron in a certain atom is equal to 3.464ħ. Which of the following angles could NOT be the angle between the orbital angular momentum vector of the electron and an arbitrary z-direction?

Accepted Answer

The orbital angular momentum vector of an electron can only be aligned parallel or antiparallel to an arbitrary z-direction, so the angle between them can only be 0° or 180°.

Question 10

Multielectron atoms: The only INVALID electron state and shell designation among the following is

Accepted Answer

The designation of 2d is not valid as there are no d orbitals in the second energy level. The correct designation for the second energy level is 2s and 2p.

Question 11

Multielectron atoms: An atom has completely filled inner shells and a single valence electron in an excited p state. The filled inner shells have an orbital momentum equal to zero. What is the magnitude of the orbital angular momentum L of the atom?

Accepted Answer

The magnitude of the orbital angular momentum of an electron in an excited p state is given by L = √(l(l+1)ħ), where l is the azimuthal quantum number. In this case, the valence electron is in an excited p state, which means l=1. Plugging this into the equation, we get L = √(1(1+1)ħ) = √2ħ, or approximately 1.4ħ. Therefore, the best choice is C.

Question 12

Multielectron atoms: The correct ground state electron configuration of boron, which has 5 electrons, is

Accepted Answer

The correct ground state electron configuration for boron, which has 5 electrons, is represented by "1s^2 2s^2 2p^1", which matches the format in choice A when considering the unusual notation provided.

Question 13

Multielectron atoms: A neutral atom has an electron configuration of 1s²2s²2p⁶3s²3p². How many protons does it have in its nucleus?

Accepted Answer

The number of protons in an atom is equal to its atomic number, which can be determined by counting the number of electrons in the neutral atom's electron configuration. In this case, there are a total of 14 electrons, so the atom has 14 protons in its nucleus.

Question 14

Angular momentum: The magnitude of the orbital angular momentum L of an electron in a certain atom is equal to 3.464ħ. Which one of the following numbers could be the principal quantum number n of the electron?

Accepted Answer

The magnitude of the angular momentum is given by L = √(l(l+1)ħ), where l is the azimuthal quantum number. Since L = 3.464ħ, we have 3.464 = √(l(l+1)), which implies l(l+1) = 12. The possible values of l are obtained by solving this quadratic equation, which gives l = 3 or -4. However, the allowed values of l are non-negative integers such that 0 ≤ l ≤ n-1. Therefore, the only possible value of n is 4, which gives l = 3.

Question 15

Multielectron atoms: How many electrons does it take to fill the d subshell?

Accepted Answer

The d subshell can hold a maximum of 10 electrons. The electron configuration for a multielectron atom fills the s and p subshells first, and then the d subshell. Therefore, it would take 10 electrons to fill the d subshell.

Question 16

Multielectron atoms: How many possible sets of quantum numbers (electron states) are there in the 5f subshell?

Accepted Answer

The 5f subshell has a maximum of 14 electrons. According to the Pauli exclusion principle, no two electrons in an atom can have the same set of four quantum numbers. Therefore, each electron in the 5f subshell can have a unique set of quantum numbers. The possible sets of quantum numbers can be calculated using the formulas for n, l, ml, and ms:

n = 5
l can range from 0 to 4 (l = 0 for s subshell, l = 1 for p subshell, l = 2 for d subshell, l = 3 for f subshell, and l = 4 for g subshell)
ml can range from -l to l
ms can be either +1/2 or -1/2

Using these formulas, we can calculate the possible sets of quantum numbers for each electron in the 5f subshell. Since there are 14 electrons in the subshell, there are a total of 14 possible sets of quantum numbers. Therefore, the answer is E) 14.

Question 17

Angular momentum: What is the greatest magnitude of the orbital angular momentum L for an electron in a state with principal quantum number

Accepted Answer

The answer of Angular momentum: What is the greatest magnitude...

Question 18

Angular momentum: What is the greatest total angular momentum J for an electron in the n = 2 shell?

Accepted Answer

The answer of Angular momentum: What is the greatest total...

Question 19

Hydrogen atom: A muonic hydrogen atom is a proton orbited by a muon (a particle with the same charge as an electron and 207 times its mass) in which the mass of the muon is significant relative to the mass of the proton. What would be the radius of the smallest muon orbit in a muonic hydrogen atom?

Accepted Answer

The radius of the smallest muon orbit can be calculated using the formula for the Bohr radius:

a = ε₀h²/(πμe²)

where ε₀ is the electric constant, h is Planck's constant, μ is the reduced mass of the system (muon-proton), and e is the elementary charge.

For a muonic hydrogen atom, the reduced mass is:

μ = m_pm_μ/(m_p + m_μ) = m_p/208

where m_p is the mass of the proton and m_μ is the mass of the muon.

Substituting in the values and solving, we get:

a = 2.84 × 10^-13 m

Therefore, the answer is D.

Question 20

Hydrogen atom: Doubly-ionized lithium is a hydrogen-like ion, in which the nucleus with charge 3e is orbited by a single electron. What wavelength light would be emitted from Li⁺⁺ as the electron dropped from the n = 4 to the n = 3 orbit?

Accepted Answer

The Rydberg formula for the wavelength of light emitted by a hydrogen-like ion is:$$\frac{1}{\lambda} = R_H * (\frac{1}{n_f^2} - \frac{1}{n_i^2})$$where:* $$\lambda$$ is the wavelength of light emitted* $$R_H$$ is the Rydberg constant for hydrogen ($$1.0973731	imes10^7 \ m^{-1}$$)* $$n_f$$ is the final energy level of the electron* $$n_i$$ is the initial energy level of the electronFor doubly-ionized lithium, $$Z = 3$$, so the Rydberg constant is:$$R_{Li^{++}} = \frac{R_H}{Z^2} = \frac{1.0973731	imes10^7 \ m^{-1}}{3^2} = 1.226	imes10^6 \ m^{-1}$$Substituting the given values into the Rydberg formula, we get:$$\frac{1}{\lambda} = 1.226	imes10^6 \ m^{-1} * (\frac{1}{3^2} - \frac{1}{4^2}) = 4.80	imes10^4 \ m^{-1}$$Therefore, the wavelength of light emitted is:$$\lambda = \frac{1}{4.80	imes10^4 \ m^{-1}} = 208 \ nm$$

Quiz 32: Electromagnetic Waves Optics