Question 1

In the subshell of the Li²⁺ ion with orbital quantum number , the allowed values of the magnetic quantum number are A) − to B) −( + 1)to( + l) C) − ( + 2) to ( + 2) D) −( + 3) to ( + 3) E) 0 to n − 1

Accepted Answer

A

Question 2

Suppose Bohr had chosen the potential energy of the electron in the hydrogen atom to be V = 0 when the electron is in the orbit with n = 1. He could do this by
A) choosing n = 1 for the orbit where the kinetic energy of the electron is zero.
B) adding a constant 13.6 eV to the potential energy for all values of n.
C) adding a constant 27.2 eV to the potential energy for all values of n.
D) subtracting a constant 13.6 eV from the potential energy for all values of n.
E) subtracting a constant 27.2 eV from the potential energy for all values of n.

Accepted Answer

C

Question 3

The energy needed to remove an electron from the first excited state of a Li²⁺ ion is A) 53 eV B) 31 eV C) 92 eV D) 122 eV E) 61 eV

Accepted Answer

The energy required to remove an electron from the first excited state of a Li²⁺ ion can be calculated using the formula for the energy levels of hydrogen-like ions: E = -13.6 * Z² / n² eV, where Z is the atomic number (3 for lithium) and n is the principal quantum number. For the first excited state, n = 2. The energy difference between n = 2 and n = ∞ (ionization) is 13.6 * 3² / 2² = 30.6 eV, approximately 31 eV.

Question 4

In a shell of the hydrogen atom with n = 3, the permitted values of the orbital magnetic quantum number  are
A) −1, 0, 1
B) 2, 1, 0
C) 2, 1, 0, −1, −2
D) 0
E) 3, 2, 1, 0, −1, −2, −3

Accepted Answer

The answer of In a shell of the hydrogen atom...

Question 5

Of the following states, 5s, 3p, 4f, 5p, 4g, 3d, and 2p, the one which is NOT allowed is
A) 3p
B) 4f
C) 3d
D) 4g
E) 2p

Accepted Answer

The quantum numbers for atomic orbitals include the principal quantum number (n), the azimuthal (angular momentum) quantum number (l), the magnetic quantum number (m_l), and the spin quantum number (m_s). The azimuthal quantum number (l) can have values from 0 to n-1. For any given value of n, l ranges from 0 to n-1, where l=0 corresponds to s orbitals, l=1 to p orbitals, l=2 to d orbitals, and l=3 to f orbitals. The "g" orbital would correspond to l=4, which is not possible for n=4 because the maximum value of l for n=4 is 3 (f orbital). Therefore, the 4g state is not allowed according to quantum mechanics.

Question 6

The allowed values of for the n = 3 shell in a Li²⁺ ion are A) 1, 2 B) 0, 1 C) 0, 1, 2 D) 0, 1, 2, 3 E) 1, 2, 3

Accepted Answer

The allowed values for the third shell (n = 3) are given by the formula 2n^2. For n = 3, this gives a maximum value of 18. However, the LiS1U1P12+ ion only has two electrons in the 3rd shell. These electrons will occupy the lowest energy levels available, which are the 3S and 3P orbitals. The 3S orbital has one allowed value of ml, which is 0. The 3P orbital has three allowed values of ml, which are -1, 0, and 1. Therefore, the allowed values of 11eb6adc_ed92_4ed1_b0cf_1db84104f3d8_TB8245_11 for the n = 3 shell in a LiS1U1P12+S1S1P0 ion are 0, 1, and 2. Hence, the answer is C.

Question 7

The K, L, M symbols represent values of the quantum number A) n B) C) D) m_s E) m_j

Accepted Answer

The K, L, M symbols are used to represent the values of the quantum number n, which indicates the energy level of an electron in an atom.

Question 8

What wavelength (in μm) is associated with the Paschen series for n = 4? (R_H = 1.097 × 10⁷ m−¹) A) 320 B) 530 C) 2.7 D) 1.9 E) 0.5

Accepted Answer

The Paschen series corresponds to transitions where the final energy level is n=3. For n=4 to n=3 transition, using the Rydberg formula, the wavelength is approximately 1.9 μm.

Question 9

The allowed values of n for the Li²⁺ ion are A) 1 to ∞ B) 2 to ∞ C) 3 to ∞ D) any real number E) 1 to 10

Accepted Answer

The allowed values of $n$ for any ion or atom in quantum mechanics, which determines the principal quantum number and thus the energy levels an electron can occupy, start from 1 and theoretically extend to infinity. This is because $n$ represents the shell level in an atom, starting from the closest to the nucleus at $n=1$ and moving outward.

Question 10

An electron is moving at a speed of 2.1 × 10⁶ m/s in the first Bohr orbit. Determine its de Broglie wavelength. A) 0.30 × 10−¹⁰ m B) 1.7 × 10−¹⁰ m C) 0.50 × 10−¹⁰ m D) 3.5 × 10−¹⁰ m E) 1.5 × 10−¹⁰ m

Accepted Answer

The answer of An electron is moving at a speed...

Question 11

How fast is the electron moving in the first Bohr orbit? A) 3.3 × 10⁶ m/s B) 2.2 × 10⁶ m/s C) 4.4 × 10⁶ m/s D) 5.5 × 10⁶ m/s E) 5.5 × 10¹⁵ m/s

Accepted Answer

The speed of electron in the first Bohr orbit can be calculated by using the formula v = (Z/n) * (e^2/4πε0m), where Z is the atomic number, n is the principal quantum number, e is the elementary charge, ε0 is the permittivity of free space and m is the mass of the electron. For hydrogen atom, Z = 1 and n = 1. Substituting the values, we get v = (1/1) * (ke^2/me^2) = (ke^2/me^2) = (9×10^9) * (1.6×10^-19)^2 / (9.1×10^-31) = 2.2 × 10^6 m/s (approximately). Therefore, the correct choice is B.

Question 12

An energy of 13.6 eV is needed to ionize an electron from the ground state of a hydrogen atom. Selecting the longest wavelength that will work from the those given below, what wavelength is needed if a photon accomplishes this task?
A) 60 nm
B) 80 nm
C) 70 nm
D) 90 nm
E) 40 nm

Accepted Answer

The energy of a photon is inversely proportional to its wavelength, as given by the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. To ionize an electron with the least amount of energy (13.6 eV), we need the longest wavelength, because longer wavelengths correspond to lower energies. Among the given options, 90 nm is the longest wavelength, making it the correct choice for achieving ionization with the minimum required energy.

Question 13

What value of wavelength is associated with the Lyman series for n = 2? (R_H = 1.097 × 10⁷ m−¹) A) 8.2 × 10⁶ m B) 1.2 × 10−⁷ m C) 2.7 × 10⁶ m D) 3.6 × 10−⁷ m E) 8.8 × 10−⁷ m

Accepted Answer

The answer of What value of wavelength is associated with...

Question 14

An electron in a hydrogen atom makes a transition from the n = 4 to the n = 3 energy state. Determine the energy (in eV) of the emitted photon.
A) 0.54
B) 0.66
C) 0.85
D) 1.51
E) 10.2

Accepted Answer

The energy difference between two energy levels in a hydrogen atom can be calculated using the formula: $$E = -13.6\,eV \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ where $n_i$ and $n_f$ are the initial and final principal quantum numbers, respectively. Plugging in the values for the transition from $n = 4$ to $n = 3$, we get: $$E = -13.6\,eV \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = -13.6\,eV \left( \frac{1}{9} - \frac{1}{16} \right)$$ $$E = -13.6\,eV \left( \frac{16 - 9}{144} \right) = -13.6\,eV \left( \frac{7}{144} \right) \approx -0.66\,eV$$ Since the energy is emitted, we take the absolute value, giving 0.66 eV.

Question 15

An electron in a hydrogen atom makes a transition from the n = 3 to the n = 1 energy state. Determine the wavelength of the emitted photon (in nm).
A) 1006
B) 209
C) 306
D) 103
E) 821

Accepted Answer

The wavelength of the emitted photon can be calculated using the Rydberg formula for hydrogen: 1/λ = R_H (1/n1² - 1/n2²), where R_H is the Rydberg constant (1.097 x 10^7 m^-1), n1 = 1, and n2 = 3. Solving this gives λ ≈ 103 nm.

Question 16

One of the main problems with the Bohr model of the hydrogen atom when compared with the results of the methods of quantum mechanics used to describe atoms, was that the Bohr model predicted A) the ground state angular momentum was L = 1 . B) the frequency of the radiation emitted when an electron "jumps" from one allowed orbit to another was hf = E_i − E_f. C) the potential energy function for the hydrogen atom was given by V(r) = −ke²/r. D) the energy of the ground state of the hydrogen atom was E_n = −13.6 eV.

Accepted Answer

The Bohr model predicted that the ground state angular momentum was L = 1, while the methods of quantum mechanics showed that the ground state of the hydrogen atom had an angular momentum of L = 0.

Question 17

The s, p, d, f, symbols represent values of the quantum number A) m_s B) n C) D) E) m_j

Accepted Answer

The answer of The s, p, d, f, symbols represent...

Question 18

A hydrogen atom is in its first excited state (n = 2). The linear momentum of the electron is (in kg ⋅ m/s) A) 3 × 10−²⁴ B) 2 × 10−²⁴ C) 1 × 10−²⁴ D) 4 × 10−²⁴ E) 3 × 10−¹⁵

Accepted Answer

The answer of A hydrogen atom is in its first...

Question 19

The number of states in the He⁺ ion corresponding to the principle quantum number n = 5 are A) 18 B) 25 C) 50 D) 9 E) 11

Accepted Answer

For the HeS1U1P1 term, there are (2S+1)(2L+1) states where S and L are the total spin and orbital angular momentum quantum numbers, respectively. For n=5, L can range from 0 to 4 (since L<n), and there are 2S+1=2 states for each value of L. So for HeS1U1P1, there are (2+1)(2(0)+1)+(2+1)(2(1)+1)+(2+1)(2(2)+1)+(2+1)(2(3)+1)+(2+1)(2(4)+1) = 3+15+35+63+99 = 215 states.

For the S1S1P0 term, the parity is even (since L=0 for P0), so there is only one state.

Therefore, the total number of states is 215*1 = 215, which corresponds to choice C.

Question 20

Light is emitted by hydrogen atoms in the visible range for a hydrogen atom. Its wavelength is 656 nm. What value of n is associated with the light? (R_H = 1.097 × 10⁷ m−¹) A) 5 B) 2 C) 4 D) 3 E) 6

Accepted Answer

The wavelength of the emitted light is related to the energy difference between the levels of the hydrogen atom. Using the Rydberg formula, we can calculate the value of n for the emitted light:

1/λ = R(1/n₁² - 1/n₂²)

where λ is the wavelength of the light, R is the Rydberg constant (1.097 × 10^7 m^-1), and n₁ and n₂ are the initial and final energy levels of the electron.

In this case, the wavelength is given as 656 nm, which we convert to meters (656 × 10^-9 m). The initial level is the ground state (n₁ = 1) and the final level is the first excited state (n₂ = 2). Substituting these values into the equation and solving for n₂:

1/656 × 10^-9 = 1.097 × 10^7 (1/1² - 1/n₂²)

1/n₂² = 1 - (656 × 10^-9)(1.097 × 10^7)

n₂ = 3

Therefore, the answer is choice D.

Quiz 42: Atomic Physics

In the subshell of the Li²⁺ ion with orbital quantum number , the allowed values of the magnetic quantum number are

Suppose Bohr had chosen the potential energy of the electron in the hydrogen atom to be V = 0 when the electron is in the orbit with n = 1. He could do this by

The energy needed to remove an electron from the first excited state of a Li²⁺ ion is

In a shell of the hydrogen atom with n = 3, the permitted values of the orbital magnetic quantum number are

Of the following states, 5s, 3p, 4f, 5p, 4g, 3d, and 2p, the one which is NOT allowed is

The allowed values of for the n = 3 shell in a Li²⁺ ion are

The K, L, M symbols represent values of the quantum number

What wavelength (in μm) is associated with the Paschen series for n = 4? (R_H = 1.097 × 10⁷ m−¹)

The allowed values of n for the Li²⁺ ion are

An electron is moving at a speed of 2.1 × 10⁶ m/s in the first Bohr orbit. Determine its de Broglie wavelength.

How fast is the electron moving in the first Bohr orbit?

An energy of 13.6 eV is needed to ionize an electron from the ground state of a hydrogen atom. Selecting the longest wavelength that will work from the those given below, what wavelength is needed if a photon accomplishes this task?

What value of wavelength is associated with the Lyman series for n = 2? (R_H = 1.097 × 10⁷ m−¹)

An electron in a hydrogen atom makes a transition from the n = 4 to the n = 3 energy state. Determine the energy (in eV) of the emitted photon.

An electron in a hydrogen atom makes a transition from the n = 3 to the n = 1 energy state. Determine the wavelength of the emitted photon (in nm) .

One of the main problems with the Bohr model of the hydrogen atom when compared with the results of the methods of quantum mechanics used to describe atoms, was that the Bohr model predicted

The s, p, d, f, symbols represent values of the quantum number

A hydrogen atom is in its first excited state (n = 2) . The linear momentum of the electron is (in kg ⋅ m/s)

The number of states in the He⁺ ion corresponding to the principle quantum number n = 5 are

Light is emitted by hydrogen atoms in the visible range for a hydrogen atom. Its wavelength is 656 nm. What value of n is associated with the light? (R_H = 1.097 × 10⁷ m−¹)