Calculating the quantum state of an electron

In summary, the electron fell from the 280th quantum state, which has a wavelength that corresponds to the same energy level as the free electron.
  • #1
bbbl67
212
21

Homework Statement


An electron in a hydrogen atom falling from an excited state to the ground state has the same wavelength than an electron moving at a speed of 7821 ms^-1. From which excited state did the electron fall from?

Homework Equations


I used the kinetic energy equation:
K = (m v^2)/2
m = 1 m_e = 9.109383×10^-31 kg
v = 7821 m/s

I also used the Rydberg Energy equation:
E_n = -h c R_inf Z^2 / n^2 |
E_n | electron energy
h | Planck constant
c | speed of light
R_inf | Rydberg constant at infinity
Z | atomic number
n | quantum number

Z = 1 (since it's hydrogen)
n = quantum number = ? (this is what we are trying to find)

The constant (h c R_inf) is also known as the Rydberg unit of energy, Ry:
Ry = h c R_inf = 13.60569 eV

The Attempt at a Solution


So this was in somebody's chemistry textbook, and I thought I'd try to solve this for giggles. I got no chemistry background, except in high school. I knew the kinetic energy equation by heart, I had to google the Rydberg equation (https://is.gd/QhBP9Z). I wonder if my logic is correct here?

kinetic energy:
K = 2.786×10^-23 J = 1.739×10^-4 eV

quantum energy:
E_n = - Ry Z^2 / n^2

Since we're only looking for the magnitude in energy differences, I think we can ignore the negative sign in the equation, and simply write it as:
E_n = Ry Z^2 / n^2

So we set,
E_n = K
K = Ry Z^2 / n^2
n^2 = Ry Z^2 / K
n = sqrt(Ry Z^2 / K)
= sqrt(13.60569 eV * 1^2 / 1.739×10^-4 eV)
= 279.7
~ 280

Conclusion: So that would mean that the electron fell from the 280th quantum state.

How was my logic about this? Is there any other way of solving this?
 
Last edited:
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  • #2
Please use LaTeX

Your energy comes out as 0.1 meV range. Something is wrong. Typical energies in atoms are eV. I don't really understand the question

bbbl67 said:
An electron in a hydrogen atom falling from an excited state to the ground state has the same wavelength ...

So which state is meant here, the excited one or the ground-state? Also, why is there emphasis on the wavelength? You don't seem to be using this bit at all in your solution. Of course, if the question does mean that the energy of the excited atom was ##-\frac{m_e (7821 \,\mbox{m/s})^2}{2}## then the quantum number of this state will be very high. Then your answer makes sense, though I think the relationship between the kinetic energy of the electron and the full energy of the atom is slightly different (use Virial Theorem).
 
  • #3
Cryo said:
Please use LaTeX

Your energy comes out as 0.1 meV range. Something is wrong. Typical energies in atoms are eV. I don't really understand the question
Well, the electron is traveling really fast almost 8 km/s, so it has a lot of kinetic energy.

Cryo said:
So which state is meant here, the excited one or the ground-state? Also, why is there emphasis on the wavelength? You don't seem to be using this bit at all in your solution. Of course, if the question does mean that the energy of the excited atom was ##-\frac{m_e (7821 \,\mbox{m/s})^2}{2}## then the quantum number of this state will be very high. Then your answer makes sense, though I think the relationship between the kinetic energy of the electron and the full energy of the atom is slightly different (use Virial Theorem).
Well, the wavelength just corresponds to being at the same energy level. Yes, well the 280th energy level would be a pretty high energy level.
 
  • #4
The energy of the electron in the atom is not equal to the kinetic energy of the free electron. It is the wavelengths that are the same. (Strictly, the question should have said "the wavelength of the photon emitted when an electron falls...")
For the atom: you need the difference in energy between the two levels, not just the energy of one level. (The ground state energy is not zero!)
For the free electron: you need the de Broglie wavelength.
 
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Likes bbbl67
  • #5
bbbl67 said:
Well, the electron is traveling really fast almost 8 km/s, so it has a lot of kinetic energy.

Typical energies in atomic physics are eV, so relative to that your 8 km/s electron does not have a lot of energy.

bbbl67 said:
Well, the wavelength just corresponds to being at the same energy level.

?
 
  • #6
mjc123 said:
The energy of the electron in the atom is not equal to the kinetic energy of the free electron. It is the wavelengths that are the same. (Strictly, the question should have said "the wavelength of the photon emitted when an electron falls...")
For the atom: you need the difference in energy between the two levels, not just the energy of one level. (The ground state energy is not zero!)
For the free electron: you need the de Broglie wavelength.
Ah, I get it now. Used the wrong formula:

λ = h/(m v) |
λ | wavelength
m | mass
v | velocity
h | Planck constant (≈ 6.62607×10^-34 J s)

λ = 6.62607×10^-34 J s / (9.109383×10^-31 kg * 7821 m/s)
= 93 nm
(https://is.gd/lLLYYt)

1/λ = R_∞ Z^2 abs(1/n_f^2 - 1/n_i^2) |
λ | photon wavelength
n_f | principal quantum number of final state
n_i | principal quantum number of initial state
Z | atomic number
R_∞ | Rydberg constant (≈ 1.0973731569×10^7 m^(-1))

Z = 1
n_i = 1
λ = 93
n_f = ?
= 7
(https://is.gd/WWIzB7)

Does this make more sense?
 

Related to Calculating the quantum state of an electron

1. What is the quantum state of an electron?

The quantum state of an electron refers to its unique set of quantum numbers, which describe its energy level, orbital shape, orientation, and spin. This state determines the electron's behavior and properties in an atom.

2. How is the quantum state of an electron calculated?

The quantum state of an electron is calculated using the Schrödinger equation, which takes into account the electron's energy and potential energy within an atom. It is a complex mathematical equation that involves solving for the wave function of the electron.

3. What is meant by the term "quantum numbers"?

Quantum numbers are a set of four values that describe the quantum state of an electron. They include the principal quantum number, azimuthal quantum number, magnetic quantum number, and spin quantum number.

4. How does the quantum state of an electron affect its behavior?

The quantum state of an electron determines how it behaves in an atom, such as its position, energy, and spin. It also determines its interactions with other electrons and atoms, and contributes to the overall properties of a material.

5. Can the quantum state of an electron change?

Yes, the quantum state of an electron can change through processes such as absorption or emission of energy, or through interactions with other particles. These changes can result in different energy levels, orbital shapes, and orientations for the electron.

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