Question 1

If light excites atomic hydrogen from its lowest energy level to the n = 12 level,what is the energy of the photons of this light?

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Question 2

What is the energy (in electron-volts)of the n = 3 state of atomic hydrogen?

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Question 3

A hydrogen atom makes a transition from the n = 8 to the n = 3 state.The wavelength of the emitted photon is closest to which one of the following values? (c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The energy difference between two levels in a hydrogen atom is given by the formula: $$E = \frac{hcR}{n_2^2} - \frac{hcR}{n_1^2}$$, where $R$ is the Rydberg constant ($1.097 \times 10^7 m^{-1}$), $h$ is Planck's constant, $c$ is the speed of light, $n_1$ and $n_2$ are the principal quantum numbers of the two levels. The wavelength of the emitted photon can be found from the energy difference using the relation $E = \frac{hc}{\lambda}$, where $\lambda$ is the wavelength. Plugging in the values for the transition from $n = 8$ to $n = 3$, and solving for $\lambda$, we find that the wavelength is closest to $9.57 \times 10^{-7} m$, which corresponds to choice A.

Question 4

The longest wavelength photon that can be emitted by a hydrogen atom,for which the final state is $n = 9 ,$ is closest to which one of the following values? (c = 3.0 × 10⁸ m/s, $1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J}$ $\left. h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$

Accepted Answer

The longest wavelength photon corresponds to the smallest energy difference between the levels involved. The energy difference between levels $n_i$ and $n_f$ is given by:$$\Delta E = E_i - E_f = -13.6 \text{ eV} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$For the longest wavelength photon, $n_i = \infty$ and $n_f = 9$. Plugging these values into the equation, we get:$$\Delta E = -13.6 \text{ eV} \left(\frac{1}{9^2} - \frac{1}{\infty^2}\right) = -13.6 \text{ eV} \left(\frac{1}{81}\right) = 0.168 \text{ eV}$$Using the relationship between energy and wavelength:$$\lambda = \frac{hc}{\Delta E}$$we get:$$\lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.0 \times 10^8 \text{ m/s})}{0.168 \times 1.60 \times 10^{-19} \text{ J}} = 39,000 \text{ nm}$$Therefore, the longest wavelength photon that can be emitted by a hydrogen atom, for which the final state is $n = 9$, is closest to 39,000 nm.

Question 5

A hydrogen atom is excited to the n = 9 level.Its decay to the n = 6 level detected in a photographic plate.What is the frequency of the light photographed? (1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The energy difference between the n = 9 and n = 6 levels can be calculated using the formula:
ΔE = E₉ - E₆ = (-(13.6 eV) / 9²) - (-(13.6 eV) / 6²) = -1.358 eV

Converting this energy difference to frequency using the formula E = hν, where h is Planck's constant:
-1.358 eV = -(1.358 × 1.60 × 10⁻¹⁹ J) = -2.1736 × 10⁻¹⁹ J
ν = E / h = (-2.1736 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J ∙ s) = 3.28 × 10¹³ Hz
Rounding to three significant figures, the frequency of the light is 5.08 × 10¹³ Hz, which corresponds to choice letter A.

Question 6

Given that the binding energy of the hydrogen atom in its ground state is -13.6 eV,what is the energy when it is in the n = 5 state?

Accepted Answer

The energy levels of a hydrogen atom can be calculated using the formula $E_n = \frac{E_1}{n^2}$, where $E_1$ is the energy of the ground state (n=1), and $n$ is the principal quantum number of the state in question. Given $E_1 = -13.6$ eV for the ground state, the energy for the $n = 5$ state is $E_5 = \frac{-13.6 \text{ eV}}{5^2} = \frac{-13.6 \text{ eV}}{25} = -0.544 \text{ eV}$.

Question 7

A hydrogen atom makes a transition from the $n = 15$ state to the $n = 5$ state.Find the wavelength of the emitted photon.(c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The energy of a hydrogen atom in the nth state is given by:$$E_n = -13.6 \text{ eV} \left(\frac{1}{n^2}\right)$$The energy difference between the $n = 15$ and $n = 5$ states is:$$\Delta E = E_5 - E_{15} = -13.6 \text{ eV} \left(\frac{1}{5^2} - \frac{1}{15^2}\right) = 1.89 \text{ eV}$$The wavelength of the emitted photon is given by:$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{1.89 \text{ eV}(1.60 \times 10^{-19} \text{ J/eV})} = 2.56 \mu \text{m}$$

Question 8

In a transition from one vibrational state to another,a molecule emits a photon of wavelength 5.56 µm.What is the energy difference between these two states? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J)

Accepted Answer

The energy of a photon 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 of the photon.

We can rearrange this formula to solve for the energy difference between two vibrational states:

ΔE = hc/λ2 - hc/λ1

where λ1 and λ2 are the wavelengths corresponding to the initial and final vibrational states, respectively.

Substituting the given values, we get:

ΔE = (6.626 × 10^-34 J ∙ s)(3.00 × 10^8 m/s)/((5.56 × 10^-6 m)^2) - (6.626 × 10^-34 J ∙ s)(3.00 × 10^8 m/s)/((λ)^2)

Simplifying this expression, we get:

ΔE = 2.2479 × 10^-19 J

Converting to electron volts, we get:

ΔE = 2.2479 × 10^-19 J / (1.60 × 10^-19 J/eV) = 0.223 eV

Therefore, the answer is choice A (0.223 eV).

Question 9

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? (1 eV = 1.60 × 10^-19 J)

Accepted Answer

CThe energy difference between the hyperfine levels can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s), $c$ is the speed of light ($3.00 \times 10^8$ m/s), and $\lambda$ is the wavelength (21 cm or 0.21 m). Substituting the values, we get $E = \frac{(6.626 \times 10^{-34} \text{ J·s})(3.00 \times 10^8 \text{ m/s})}{0.21 \text{ m}}$, which gives $E \approx 9.46 \times 10^{-25}$ J. To convert this energy to electron volts (eV), we use the conversion factor $1 \text{ eV} = 1.60 \times 10^{-19}$ J, resulting in $E \approx 5.9 \times 10^{-6}$ eV. Therefore, the correct answers are the ones that correspond to these values in eV and J.

Question 10

A hydrogen atom is excited to the n = 11 level.Its decay to the n = 7 level is detected in a photographic plate.What is the wavelength of the light photographed? (c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The energy difference between the n = 11 and n = 7 levels can be calculated using the Balmer-Rydberg equation:
$\frac{1}{\lambda}=R_H(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}})$
where $R_H=1.096776	imes10^7 m^{-1}$ is the Rydberg constant for hydrogen, $n_1=7$, and $n_2=11$. Solving for $\lambda$, we get:
$\lambda= 7510$ nm. 
Therefore, the correct answer is choice (A).

Question 11

What is the shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is n = 3? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J)

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Question 12

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 which one of the following values? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J)

Accepted Answer

The longest wavelength corresponds to the lowest energy transition, which for a hydrogen atom initially in the n = 3 state is the transition to n = 2. The energy of this transition can be calculated using the Rydberg formula for hydrogen: $\Delta E = 13.6 \text{ eV} \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$. This gives $\Delta E \approx 1.89 \text{ eV}$. Converting this energy to joules ($1.89 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV}$) and then using the relation $E = \frac{hc}{\lambda}$ to find the wavelength $\lambda$, we find that $\lambda$ is closest to 658 nm.

Question 13

If a hydrogen atom in the ground state absorbs a photon of energy 12.09 eV,to which energy state will it make a transition?

Accepted Answer

The photon's energy of 12.09 eV is equal to the energy difference between the ground state (n=1) and the first excited state (n=2). Therefore, the hydrogen atom will make a transition to the n=2 energy level.

Question 14

Light shines through atomic hydrogen gas that was initially in its ground state.You observe that after awhile much of the hydrogen gas has been excited to its n = 5 state.What wavelength of light entering the gas caused this excitation? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J)

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Question 15

Consider the Bohr model for the hydrogen atom.(c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s) (a)How much energy (in eV)is needed to cause a transition of the atom from the second excited state to the third excited state? (b)What wavelength photon just has enough energy to initiate the transition in (a)?

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Question 16

What is the ionization energy of the neutral hydrogen atom?

Accepted Answer

The ionization energy of the neutral hydrogen atom is 13.6 eV. This means that it requires 13.6 eV of energy to remove the electron from the hydrogen atom and form a hydrogen ion.

Question 17

What is the energy required to remove the electron from a hydrogen atom in the n = 11 state?

Accepted Answer

The energy required to remove an electron from a hydrogen atom in a specific quantum state (n) can be calculated using the formula E = 13.6 eV * (1/n^2), where 13.6 eV is the ionization energy of a hydrogen atom in its ground state (n=1). For n=11, E = 13.6 eV * (1/11^2) = 13.6 eV * (1/121) ≈ 0.112 eV.

Question 18

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 which one of the following values? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \right.$ $h = 6.626 \times 10 ^{- 34} \mathrm {~J} \cdot \mathrm { s }$ $1 \mathrm { eV } = 1.60 \times 10^{ - 19} \mathrm {~J} )$

Accepted Answer

The shortest wavelength corresponds to the largest energy difference between the levels, which is for $n_f = 1$. The energy difference is:$$\Delta E = E_f - E_i = -13.6 \left(\frac{1}{1^2} - \frac{1}{12^2}\right) \mathrm{eV} = 10.2 \mathrm{eV}$$The wavelength is then:$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s})(3.0 \times 10^8 \mathrm{m/s})}{10.2 \mathrm{eV}(1.60 \times 10^{-19} \mathrm{J/eV})} = 1.21 \times 10^{-7} \mathrm{m} = 121 \mathrm{nm}$$The closest choice is 92 nm.

Question 19

Given that the energy states of the hydrogen atom are given by E_n = $\frac { \text { Rch } } { n ^ { 2 } }$ ,where R = 1.097 × 10⁷ m^-1,what wavelength photon is emitted when the atom undergoes a transition from the n = 4 to the n = 6 state?

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Question 20

What is the energy of the photon emitted when a hydrogen atom undergoes a transition from the $n = 20$ state to the $n = 7$ state?

Accepted Answer

The energy of a photon is given by the equation E = hf, where h is Planck's constant and f is the frequency of the photon. The frequency of a photon is given by the equation f = c/\lambda, where c is the speed of light and \lambda is the wavelength of the photon. The wavelength of a photon is given by the equation \lambda = hc/E, where hc is Planck's constant times the speed of light.The energy of the photon emitted when a hydrogen atom undergoes a transition from the $n = 20$ state to the $n = 7$ state is given by the equation:$$E = hf = hc/\lambda = hc/(hc/E) = E$$$$E = (13.6 \text{ eV})[(1/7^2) - (1/20^2)] = 0.244 \text{ eV}$$Therefore, the correct answer is A) 0.244 eV.

Quiz 28: Atomic Physics

If light excites atomic hydrogen from its lowest energy level to the n = 12 level,what is the energy of the photons of this light?

What is the energy (in electron-volts)of the n = 3 state of atomic hydrogen?

A hydrogen atom makes a transition from the n = 8 to the n = 3 state.The wavelength of the emitted photon is closest to which one of the following values? (c = 3.00 × 108 m/s,1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

A hydrogen atom is excited to the n = 9 level.Its decay to the n = 6 level detected in a photographic plate.What is the frequency of the light photographed? (1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

Given that the binding energy of the hydrogen atom in its ground state is -13.6 eV,what is the energy when it is in the n = 5 state?

A hydrogen atom makes a transition from the n=15n = 15n=15 state to the n=5n = 5n=5 state.Find the wavelength of the emitted photon.(c = 3.00 × 108 m/s,1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

In a transition from one vibrational state to another,a molecule emits a photon of wavelength 5.56 µm.What is the energy difference between these two states? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

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? (1 eV = 1.60 × 10-19 J)

A hydrogen atom is excited to the n = 11 level.Its decay to the n = 7 level is detected in a photographic plate.What is the wavelength of the light photographed? (c = 3.00 × 108 m/s,1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

What is the shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is n = 3? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

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 which one of the following values? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)