Question 1

A compact disc of a CD player has a moment of inertia I = 2.5 × 10^-5 kg · m² and rotates at 500 rev/min. Taking L = Iω and using the quantization of angular momentum, the approximate value of n is = 1.055 × 10^-34 J · s) A) 1.2 × 10³². B) 2.0 × 10³⁰ ^. C) 1.2 × 10³¹. D) 130. E) 3.5 × 10¹⁵.

Accepted Answer

The answer of A compact disc of a CD player...

Question 2

The set of quantum numbers for the probability density shown is
A) n = 2,  = 0, m = 0.
B) n = 2,  = 0, m = 1.
C) n = 2,  = 1, m = 0.
D) n = 1,  = 0, m = 0.
E) n = 1,  = 1, m = 0.

Accepted Answer

The answer of  The set of quantum numbers for...

Question 3

Consider a beryllium z = 4) ion with all but one of its electrons removed. What is the kinetic energy of the electron in the n = 1 orbit. The radius of the first Bohr orbit in Be³⁺ is 0.0132 nm.) A) 54.4 eV B) 218 eV C) 13.6 eV D) 27.2 eV E) 109 eV

Accepted Answer

The kinetic energy of the electron in the first Bohr orbit can be calculated using the formula: $K.E. = \frac{1}{2}mv^2 = \frac{1}{2}\frac{Ze^2}{4\pi\epsilon_0r}$, where $m$ is the mass of the electron, $v$ is its velocity, $Z$ is the nuclear charge, $e$ is the elementary charge, $\epsilon_0$ is the permittivity of free space, and $r$ is the radius of the first Bohr orbit.

Substituting the given values, we get:

$K.E. = \frac{1}{2}\frac{(4)(1.602	imes10^{-19}	ext{ C})^2}{4\pi(8.854	imes10^{-12}	ext{ C}^2/	ext{N}\cdot	ext{m}^2)(0.0132	imes10^{-9}	ext{ m})}$

$K.E. = 218	ext{ eV}$

Therefore, the correct answer is B.

Question 4

In the Bohr model of the hydrogen atom, what is the magnitude of the potential energy of an electron in the n = 11 orbit? The radius of the first Bohr orbit is 0.0529 nm.) A) 2.25 × 10^-1 eV B) 1.12 × 10^-1 eV C) 5.62 × 10^-2 eV D) 2.47 eV E) None of these is correct.

Accepted Answer

The potential energy (PE) in the Bohr model is given by PE = -2 * (13.6 eV) / n². For n = 11, PE = -2 * (13.6 eV) / 11² = -0.225 eV, and the magnitude is 0.225 eV, which is approximately 2.25 × 10^-1 eV.

Question 5

What is the wavelength of incident light needed for an electron to make the transition labeled f)?
A) 151 nm
B) 365 nm
C) 422 nm
D) 821 nm
E) 1459 nm

Accepted Answer

The answer of  What is the wavelength of incident...

Question 6

In the Bohr model of the hydrogen atom, what is the kinetic energy of the electron in the n = 3 orbit? The radius of the first Bohr orbit is 0.0529 nm.)
A) 3.44 eV
B) 1.51 eV
C) 3.02 eV
D) 0.75 eV
E) 4.53 eV

Accepted Answer

The kinetic energy of the electron in the nth orbit of a hydrogen atom can be found using the formula:

$E_k = \frac{-13.6	ext{ eV}}{n^2}$

For n = 3, the kinetic energy is:

$E_k = \frac{-13.6	ext{ eV}}{3^2} = -1.51	ext{ eV}$

Since kinetic energy cannot be negative, we take the absolute value:

$|E_k| = 1.51	ext{ eV}$

Therefore, the answer is B.

Question 7

For the principal quantum number n = 4, the total number of values the magnetic quantum number m can have is
A) 4.
B) 3.
C) 7.
D) 16.
E) 6.

Accepted Answer

For n = 4, the azimuthal quantum number l can be 0, 1, 2, or 3. The magnetic quantum number m ranges from -l to +l, giving 1, 3, 5, and 7 values respectively for each l. Adding these gives a total of 16 values.

Question 8

The set of quantum numbers for the probability density shown is
A) n = 2,  = 0, m = 0.
B) n = 2,  = 1, m = 0.
C) n = 2,  = 1, m = 1.
D) n = 2,  = 0, m = 1.
E) n = 1,  = 1, m = 0.

Accepted Answer

From the probability density shown, we can identify the number of nodes in the wave function as 1, indicating an l value of 1. Therefore, the correct value of 11eb1367_50f7_852b_8021_b5b68d7c9dab_TB7291_11 is 1. Additionally, we can see that the density is symmetric about the xz-plane, indicating that the m value must be 0. Finally, the wave function reaches its maximum on the xy-plane, indicating that the value of n must be 2. Therefore, the correct set of quantum numbers is n = 2, 11eb1367_50f7_852b_8021_b5b68d7c9dab_TB7291_11 = 1, m = 0.

Question 9

The Pauli exclusion principle states that no two electrons in the same atom can have the same set of quantum numbers. This set of quantum numbers is A) n, j, m_j, m_s_. B) n, l, s, j. C) n, l, m, m_s_. D) n, l, m, m_j_. E) n, p_$\theta$, p_s, m.

Accepted Answer

The correct set of quantum numbers for an electron in an atom is n (principal quantum number), l (azimuthal quantum number), m (magnetic quantum number), and m_s (spin quantum number).

Question 10

The possible orbital quantum numbers of an electron are 0, 1, 3, 4, and 5. Its principal quantum number, n, and the largest magnetic quantum number, m_l_, are A) 5 and 5. B) 6 and 5. C) 4 and 3. D) 6 and 4. E) 5 and 4.

Accepted Answer

The answer of The possible orbital quantum numbers of an...

Question 11

The symbol that represents the orbital quantum number is
A) n.
B) m.
C) L.
D) l.
E) z.

Accepted Answer

The symbol that represents the orbital quantum number is "l". "n" represents the principal quantum number, "m" represents the magnetic quantum number, "L" is not a valid quantum number symbol, and "z" is not a quantum number symbol at all.

Question 12

The symbol that represents the orbital angular momentum is
A) n.
B) m.
C) L.
D) l.
E) z.

Accepted Answer

The symbol that represents the orbital angular momentum is C (L). The quantum number 'l' describes the shape of the orbital, but the letter 'l' cannot represent the orbital angular momentum as a whole. Therefore, the correct answer would be C (L).

Question 13

The p state of an electronic configuration corresponds to
A) n = 2.
B) l = 2.
C) l = 1.
D) n = 0.
E) l = 0.

Accepted Answer

The p state in an electronic configuration corresponds to the azimuthal quantum number $l = 1$.

Question 14

The set of quantum numbers for the probability density shown is
A) n = 2,  = 0, m = 0.
B) n = 2,  = 0, m = 2.
C) n = 2,  = 1, m= ±1.
D) n = 1,  = 0, m = ±2.
E) n = 1,  = 1, m = 0 .

Accepted Answer

The probability density shown in the diagram has two lobes with opposite signs, indicating that the electron is in a p orbital. The p orbital has n = 2 and 11eb1367_50f7_ac42_8021_c9516843e49d_TB7291_11 = 1. The value of m is determined by the number of lobes, which is 2l + 1. In this case, there are two lobes, so m can be ±1. Therefore, the correct set of quantum numbers is n = 2, 11eb1367_50f7_ac42_8021_c9516843e49d_TB7291_11 = 1, m= ±1.

Question 15

The total number of distinct electron states including spin) when n = 5 is
A) 18.
B) 22.
C) 25.
D) 50.
E) 72.

Accepted Answer

The total number of distinct electron states for a given principal quantum number n is given by 2n^2. For n = 5, this equals 2*(5^2) = 2*25 = 50.

Question 16

The energy of an electron in the n = 5 state in a hydrogen atom is approximately
A) -68 eV.
B) -0.54 eV.
C) +68 eV.
D) +2.7 eV.
E) +0.54 eV.

Accepted Answer

The energy of an electron in a hydrogen atom is given by the formula E_n = -13.6 eV/n^2. For n = 5, E_5 = -13.6 eV/25 = -0.54 eV.

Question 17

If the angular momentum is characterized by the quantum number = 3, what are the possible values of L_z in units of ? A) -3, 0, +3 B) -3, -2, -1, 0 C) -2, -1, 0, +1, +2 D) -3, -2, -1, 0, +1, +2, +3

Accepted Answer

The answer of If the angular momentum is characterized by...

Question 18

For the principal quantum number n = 4, the number of values the orbital quantum number  can have is
A) 4.
B) 3.
C) 7.
D) 16.
E) 6.

Accepted Answer

The answer of For the principal quantum number n =...

Question 19

The symbol that represents the principal quantum number is
A) n.
B) m.
C) L.
D) l.
E) z.

Accepted Answer

The principal quantum number is represented by the symbol "n".

Question 20

The number of oxygen's eight electrons that are found in the p state is
A) zero.
B) 2.
C) 4.
D) 6.
E) 8.

Accepted Answer

Oxygen has 6 electrons in its p atomic orbital.

Quiz 26: Quantum Physics and Atomic Structure

A compact disc of a CD player has a moment of inertia I = 2.5 × 10^-5 kg · m² and rotates at 500 rev/min. Taking L = Iω and using the quantization of angular momentum, the approximate value of n is = 1.055 × 10^-34 J · s)

The set of quantum numbers for the probability density shown is

Consider a beryllium z = 4) ion with all but one of its electrons removed. What is the kinetic energy of the electron in the n = 1 orbit. The radius of the first Bohr orbit in Be³⁺ is 0.0132 nm.)

In the Bohr model of the hydrogen atom, what is the magnitude of the potential energy of an electron in the n = 11 orbit? The radius of the first Bohr orbit is 0.0529 nm.)

What is the wavelength of incident light needed for an electron to make the transition labeled f) ?

In the Bohr model of the hydrogen atom, what is the kinetic energy of the electron in the n = 3 orbit? The radius of the first Bohr orbit is 0.0529 nm.)

For the principal quantum number n = 4, the total number of values the magnetic quantum number m can have is

The set of quantum numbers for the probability density shown is

The Pauli exclusion principle states that no two electrons in the same atom can have the same set of quantum numbers. This set of quantum numbers is

The possible orbital quantum numbers of an electron are 0, 1, 3, 4, and 5. Its principal quantum number, n, and the largest magnetic quantum number, m_l_, are

The symbol that represents the orbital quantum number is

The symbol that represents the orbital angular momentum is

The p state of an electronic configuration corresponds to

The set of quantum numbers for the probability density shown is

The total number of distinct electron states including spin) when n = 5 is

The energy of an electron in the n = 5 state in a hydrogen atom is approximately

If the angular momentum is characterized by the quantum number = 3, what are the possible values of L_z in units of ?

For the principal quantum number n = 4, the number of values the orbital quantum number can have is

The symbol that represents the principal quantum number is

The number of oxygen's eight electrons that are found in the p state is

Quiz 26: Quantum Physics and Atomic Structure

A compact disc of a CD player has a moment of inertia I = 2.5 × 10-5 kg · m2 and rotates at 500 rev/min. Taking L = Iω and using the quantization of angular momentum, the approximate value of n is = 1.055 × 10-34 J · s)

The set of quantum numbers for the probability density shown is

Consider a beryllium z = 4) ion with all but one of its electrons removed. What is the kinetic energy of the electron in the n = 1 orbit. The radius of the first Bohr orbit in Be3+ is 0.0132 nm.)

In the Bohr model of the hydrogen atom, what is the magnitude of the potential energy of an electron in the n = 11 orbit? The radius of the first Bohr orbit is 0.0529 nm.)

What is the wavelength of incident light needed for an electron to make the transition labeled f) ?

In the Bohr model of the hydrogen atom, what is the kinetic energy of the electron in the n = 3 orbit? The radius of the first Bohr orbit is 0.0529 nm.)

For the principal quantum number n = 4, the total number of values the magnetic quantum number m can have is

The set of quantum numbers for the probability density shown is

The Pauli exclusion principle states that no two electrons in the same atom can have the same set of quantum numbers. This set of quantum numbers is

The possible orbital quantum numbers of an electron are 0, 1, 3, 4, and 5. Its principal quantum number, n, and the largest magnetic quantum number, ml, are

The symbol that represents the orbital quantum number is

The symbol that represents the orbital angular momentum is

The p state of an electronic configuration corresponds to

The set of quantum numbers for the probability density shown is

The total number of distinct electron states including spin) when n = 5 is

The energy of an electron in the n = 5 state in a hydrogen atom is approximately

If the angular momentum is characterized by the quantum number = 3, what are the possible values of Lz in units of ?

For the principal quantum number n = 4, the number of values the orbital quantum number can have is

The symbol that represents the principal quantum number is

The number of oxygen's eight electrons that are found in the p state is

A compact disc of a CD player has a moment of inertia I = 2.5 × 10^-5 kg · m² and rotates at 500 rev/min. Taking L = Iω and using the quantization of angular momentum, the approximate value of n is = 1.055 × 10^-34 J · s)

Consider a beryllium z = 4) ion with all but one of its electrons removed. What is the kinetic energy of the electron in the n = 1 orbit. The radius of the first Bohr orbit in Be³⁺ is 0.0132 nm.)

The possible orbital quantum numbers of an electron are 0, 1, 3, 4, and 5. Its principal quantum number, n, and the largest magnetic quantum number, m_l_, are

If the angular momentum is characterized by the quantum number = 3, what are the possible values of L_z in units of ?