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## 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?

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