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?
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?
K = 2.786×10^-23 J = 1.739×10^-4 eV
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)
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?