Question 1

The wavelength of a ruby laser is 694.3 nm.What is the energy difference between the two energy states for the transition that produces this light? (c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The energy difference ($E$) between the two energy states can be calculated using the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. Substituting the given values: $E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{694.3 \times 10^{-9} \text{ m}}$. Converting the result to electron volts (eV) using $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$, the calculation yields approximately 1.79 eV.

Question 2

What is the energy of the photon emitted when an electron drops from the 
State to the 
State in a hydrogen atom?

Accepted Answer

The answer of What is the energy of the photon...

Question 3

What is the wavelength of the photon emitted when an electron in a hydrogen atom which is in the initial state n= 8 jumps to the final state n = 2? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J)

Accepted Answer

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

Question 4

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?

Accepted Answer

The answer of If light excites atomic hydrogen from its...

Question 5

The longest wavelength photon that can be emitted by a hydrogen atom,for which the final state is Is closest to which one of the following values? (c = 3.0 × 10⁸ m/s,

Accepted Answer

The answer of The longest wavelength photon that can be...

Question 6

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 corresponds to the energy required to remove the electron completely from the hydrogen atom, leaving only a proton.

Question 7

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

The energy difference can be calculated using the formula E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J·s), c is the speed of light (3 × 10^8 m/s), and λ is the wavelength (21 cm = 0.21 m). Converting the result to eV gives approximately 5.9 × 10^-6 eV.

Question 8

The shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is 
Is closest to which one of the following values?

Accepted Answer

The answer of The shortest wavelength of a photon that...

Question 9

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

We can use the equation:

E = hc/λ

where E is the energy difference, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

Plugging in the given values, we get:

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

Simplifying, we get:

E = 3.57 × 10^-20 J

Finally, we can convert this to electron volts (eV) using the conversion factor given:

E = (3.57 × 10^-20 J)(1 eV/1.60 × 10^-19 J)

Simplifying again, we get:

E = 0.223 eV

Therefore, the answer is choice A.

Question 10

In a hydrogen atom,the electron 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 wavelength of the emitted photon can be calculated using the Rydberg formula for the energy difference between two hydrogen atom states, which is then converted to wavelength using the relation $E = \frac{hc}{\lambda}$. The energy levels of a hydrogen atom are given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. The energy difference between the $n = 8$ and $n = 3$ levels is calculated and then used to find the wavelength of the emitted photon. The closest value provided is $9.57 \times 10^{-7} \text{ m}$, which corresponds to the typical results of such calculations considering the energy levels involved.

Question 11

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 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. For a hydrogen atom transitioning from n=3, the lowest energy transition is to n=2 (n=3 to n=2). The energy of a photon emitted during a transition is given by the Rydberg formula: $$E = \frac{hcR_H}{n_f^2} - \frac{hcR_H}{n_i^2}$$ where $R_H$ is the Rydberg constant for hydrogen ($1.097 \times 10^7 m^{-1}$), $h$ is Planck's constant ($6.626 \times 10^{-34} J \cdot s$), $c$ is the speed of light ($3.00 \times 10^8 m/s$), $n_i$ is the initial energy level (3), and $n_f$ is the final energy level (2). Rearranging the energy equation to solve for wavelength ($\lambda$) gives: $$\lambda = \frac{hc}{E}$$ Substituting the values and solving for $\lambda$ yields a wavelength closest to 658 nm.

Question 13

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)

Accepted Answer

The energy difference between the ground state and the n = 5 state is:$$E_5 - E_1 = (-\frac{13.6}{1^2} + \frac{-13.6}{5^2}) eV = 12.09 eV$$The wavelength of light required to excite the hydrogen gas to the n = 5 state is:$$\lambda = \frac{hc}{E_5 - E_1} = \frac{(6.626 	imes 10^{-34} J \cdot s)(3.00 	imes 10^8 m/s)}{12.09 eV 	imes (1.60 	imes 10^{-19} J/eV)} = 95.2 nm$$

Question 14

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

We can use the formula for calculating the frequency of light emitted during a transition in a hydrogen atom:

v = R*(1/n1^2 - 1/n2^2)

where R is the Rydberg constant, n1 and n2 are the initial and final energy levels, respectively.

Substituting n1 = 9 and n2 = 6, and using the value of R, we get:

v = 3.29 × 10^15 Hz

This matches with option A, which is the correct answer.

Question 15

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 16

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 given state is given by the formula E = -13.6 eV/n^2. For n = 11, E = -13.6 eV/11^2 = -0.112 eV. The energy required is the positive value, 0.112 eV.

Question 17

A hydrogen atom makes a downward transition from the State to the 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 answer of A hydrogen atom makes a downward transition...

Question 18

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

Accepted Answer

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

Question 19

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)

Accepted Answer

The answer of What is the shortest wavelength of a...

Question 20

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

Accepted Answer

The energy of the absorbed photon must equal the energy difference between the initial and final states of the electron in the hydrogen atom. In this case, the energy required to move the electron from the ground state (n=1) to the first excited state (n=2) is 10.2 eV, which is less than the energy of the absorbed photon (12.09 eV). Therefore, the electron will transition to the n=2 state, making the answer B the best choice.

Quiz 31: Atomic Physics

The wavelength of a ruby laser is 694.3 nm.What is the energy difference between the two energy states for the transition that produces this light? (c = 3.00 × 108 m/s,1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

What is the energy of the photon emitted when an electron drops from the State to the State in a hydrogen atom?

What is the wavelength of the photon emitted when an electron in a hydrogen atom which is in the initial state n = 8 jumps to the final state n = 2? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

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?

The longest wavelength photon that can be emitted by a hydrogen atom,for which the final state is Is closest to which one of the following values? (c = 3.0 × 108 m/s,

What is the ionization energy of the neutral hydrogen atom?

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)

The shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is Is closest to which one of the following values?

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)

In a hydrogen atom,the electron 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 = 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)

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)

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)