Question 1

A hydrogen atom is in its n = 2 excited state when its electron absorbs 9.5 eV in an interaction with a photon.What is the energy of the resulting free electron?

Accepted Answer

A

Question 2

Light excites atomic hydrogen from its lowest level to the n = 12 level.What is the energy of the light?

Accepted Answer

A

Question 3

The electron in a hydrogen atom is infinitely distant at zero potential energy and is at an average radius of ao = 0.0529 nm when bound by 13.6 eV.A barely bound H atom may have an average radius as large as a bacterium,R = 14 μm.What is the nearest principal quantum number in this state?

Accepted Answer

The principal quantum number n is related to the radius of the electron's orbit in a hydrogen atom by the formula r = n^2 * a0. Solving for n when r = 14 μm and a0 = 0.0529 nm gives n ≈ 514.

Question 4

In the spectrum of hydrogen the lines obtained by setting m = 1 in Balmer's formula is called the Lyman series.Calculate the wavelength of the spectral line of the 16th member of the Lyman series.

Accepted Answer

The Lyman series corresponds to transitions where the final energy level is n=1. The 16th member means the initial energy level is n=17. Using the Rydberg formula for hydrogen, 1/λ = R_H (1/1^2 - 1/17^2), where R_H is the Rydberg constant (1.097 x 10^7 m^-1). Solving this gives λ ≈ 91.6 nm.

Question 5

The shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is n = 3,is closest to:

Accepted Answer

The shortest wavelength corresponds to the largest energy difference, which occurs when the electron transitions from n = 3 to n = 1. The energy difference between these levels is given by:E = -13.6 eV * (1/1^2 - 1/3^2) = -12.1 eVThe wavelength of a photon with this energy is given by:λ = hc/E = (6.626 x 10^-34 J s) * (3 x 10^8 m/s) / (-12.1 eV * 1.602 x 10^-19 J/eV) = 820 nmTherefore, the shortest wavelength of a photon that can be emitted by a hydrogen atom, for which the initial state is n = 3, is closest to 820 nm.

Question 6

The longest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is n = 3,is closest to:

Accepted Answer

The longest wavelength corresponds to the smallest energy transition. For n = 3, the smallest transition is from n = 3 to n = 2. Using the Rydberg formula, the wavelength is calculated to be approximately 656 nm, which is closest to 650 nm.

Question 7

The shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is n = 12,is closest to:

Accepted Answer

The shortest wavelength corresponds to the largest energy difference, which occurs when the electron transitions from n = 12 to n = 1. The energy difference is given by:E = -13.6 eV * (1/1^2 - 1/12^2) = 10.2 eVThe wavelength is given by:λ = hc/E = (1240 eV nm) / 10.2 eV = 121.6 nmThe closest choice is A) 92 nm.

Question 8

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.The magnitude of the orbital angular momentum of the atom is closest to:

Accepted Answer

The answer of An atom has completely filled inner shells...

Question 9

The longest wavelength of a photon that can be emitted by a hydrogen atom,for which the final state is n = 9,is closest to:

Accepted Answer

The longest wavelength involves the transition from n=10 to n=9. Using the Rydberg formula, we have:

1/λ = R(1/nf^2 - 1/ni^2)

where R is the Rydberg constant (1.097 x 10^7 m^-1), ni is the initial energy level (10 in this case), and nf is the final energy level (9). Plugging in the numbers, we get:

1/λ = 1.097 x 10^7 (1/81 - 1/100)

1/λ = 2.466 x 10^-5

λ = 40544.8 nm

Note that we rounded to four significant figures, which gives us answer A) 39,000 nm as the closest choice.

Question 10

A hydrogen atom is excited to the n = 10 level.Its decay to the n = 4 level is detected in a photographic plate.What is the frequency of the light photographed?

Accepted Answer

The frequency of the light emitted during the decay is given by the formula:
f = (E_n1 - E_n2)/h
where E_n1 and E_n2 are the energies of the initial and final states, respectively, and h is Planck's constant.
For this problem, we have:
E_n1 = -13.6 / 10^2 = -0.136 eV
E_n2 = -13.6 / 4^2 = -0.85 eV
Converting these energies to joules:
E_n1 = -2.18 x 10^-18 J
E_n2 = -1.36 x 10^-18 J
Substituting into the formula:
f = (1.36 - (-2.18)) x 10^18 / 6.626 x 10^-34
f = 1.73 x 10^15 Hz
Which corresponds to choice A) 1.73 × 10^15 Hz.

Question 11

Radio astronomers often study the radiation emitted by a hydrogen atom from a transition between the two hyperfine levels associated with the ground state.This radiation has a wavelength of 21 cm.What is the energy difference between the hyperfine levels?

Accepted Answer

The energy difference can be calculated using the formula E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. Substituting the given wavelength of 21 cm, the energy difference is approximately 5.92 × 10^-6 eV.

Question 12

The energy required to remove the electron from a hydrogen atom in the n = 11 state is closest to:

Accepted Answer

The answer of The energy required to remove the electron...

Question 13

One of the emission lines described by the original version of Balmer's formula has wavelength 372 nm.What is the value of n in Balmer's formula that gives this emission line?

Accepted Answer

The answer of One of the emission lines described by...

Question 14

Light shines through atomic hydrogen gas.It is seen that the gas absorbs light readily at a wavelength of 91.33 nm.What is the level to which the hydrogen is being excited by the absorption of light of this wavelength? Assume that the most of the atoms in the gas are in the lowest level.

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy of a photon with a wavelength of 91.33 nm is 13.6 eV. This is the energy difference between the ground state and the first excited state of hydrogen. Therefore, the hydrogen is being excited to the first excited state by the absorption of light of this wavelength.

Question 15

Find the energy of the photon emitted when an electron drops from the n = 20 state to the n = 7 state in a hydrogen atom.

Accepted Answer

The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the photon. The frequency of a photon emitted when an electron drops from a higher energy level to a lower energy level is given by f = (E_i - E_f)/h, where E_i is the initial energy of the electron and E_f is the final energy of the electron. In this case, the electron drops from the n = 20 state to the n = 7 state, so the energy of the photon emitted is given by:E = hf = (E_20 - E_7)/h = (-13.6 eV/n^2) * (1/20^2 - 1/7^2) = 0.244 eVTherefore, the correct answer is A.

Question 16

Calculate the orbital Bohr radius of the n = 3 excited state in a hydrogen atom.

Accepted Answer

The Bohr radius for the n-th excited state in a hydrogen atom is given by $ a_n = n^2 \times a_0 $, where $ a_0 $ is the Bohr radius for the ground state (approximately 0.0529 nm). For n = 3, $ a_3 = 3^2 \times 0.0529 \, \text{nm} = 0.476 \, \text{nm} $.

Question 17

A hydrogen atom makes a downward transition from the n = 20 state to the n = 5 state.Find the wavelength of the emitted photon.

Accepted Answer

The energy difference between the final and initial states can be calculated using the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength of the emitted photon, R is the Rydberg constant (1.097 × 10^7 m^-1), n1 is the initial state (n=20), and n2 is the final state (n=5).

1/λ = 1.097 × 10^7 m^-1 (1/20^2 - 1/5^2)
1/λ = 1.097 × 10^7 m^-1 (0.00225)
1/λ = 24,682.5 m^-1
λ = 4.05 × 10^-5 m
Convert to micrometers:
λ = 2.43 μm

Therefore, the correct answer is A) 2.43 μm.

Question 18

What is the wavelength of the light emitted by atomic Hydrogen according to Balmer's formula with m = 3 and n = 8?

Accepted Answer

Balmer's formula for the wavelength of light emitted by hydrogen is given by 1/λ = R_H (1/m² - 1/n²), where R_H is the Rydberg constant (1.097 x 10^7 m^-1). For m = 3 and n = 8, the calculation gives a wavelength of approximately 955 nm.

Question 19

A hydrogen atom is excited to the n = 6 level.Its decay to the n = 2 level is detected in a photographic plate.What is the wavelength of the detected emission?

Accepted Answer

The transition from n = 6 to n = 2 in a hydrogen atom corresponds to the Balmer series, which is in the visible spectrum. The wavelength for this transition is approximately 410 nm.

Question 20

What is the frequency of the light emitted by atomic Hydrogen according to Balmer's formula with m = 8 and n = 12?

Accepted Answer

Balmer's formula: 1/λ = R(1/n2 - 1/m2)where R is the Rydberg constant (1.097 × 107 m-1)1/λ = (1.097 × 107 m-1)(1/122 - 1/82)1/λ = (1.097 × 107 m-1)(1/144 - 1/64)1/λ = (1.097 × 107 m-1)(1/80)λ = 80/1.097 × 107 m-1λ = 7.29 × 10-8 mf = c/λf = (3 × 108 m/s)/(7.29 × 10-8 m)f = 2.86 × 1013 Hz

Quiz 29: Atoms and Molecules