# Calculating the quantum state of an electron

bbbl67

## 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.

Last edited:

Gold Member

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

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).

bbbl67

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.

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.

Homework Helper
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.

• bbbl67
Gold Member
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.

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

?

bbbl67
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?