Calculating Energy Levels and Wavelengths in the Hydrogen Atom

AI Thread Summary
In the hydrogen atom, when the electron transitions from the n = 1 level to the n = 4 level, a photon is absorbed. The change in energy (ΔE) during this transition is calculated as 12.75 eV, indicating an increase in energy. The transition can be described as the electron moving from a lower energy state (n = 1) to a higher energy state (n = 4). The wavelength of the absorbed photon is approximately 9.74 x 10^-8 m. The calculations are generally correct, but using hc = 1240 eV nm can simplify the wavelength calculation.
Kennedy111
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Homework Statement


In the hydrogen atom, the electron jumps from the n = 1 level to the n = 4 level.
a. During this transition from n = 1 to n = 4, is a photon emitted or absorbed?
b. What is the change in energy of the electron?
c. State the transition in words.
d. What is the wavelength of the emitted or absorbed photon?

Energy when n = 1 : -13.6 eV
Energy when n = 4 : -0.85 eV

Homework Equations


ΔE = Ef - Ei
λ = hc/E


The Attempt at a Solution


a. I believe I understand this question. I would say that the photon is absorbed...
b. ΔE = Ef - Ei
ΔE = (-0.85 eV) - (-13.6 eV)
= 12.75 eV
c. The electron has made a transition from the n = 1 to the n = 4 energy level.. is that it?? I feel like this question was too easy...
d. I used λ = hv/E but I'm not sure if that was the right formula to use...
λ = (4.14 x 10^-15 eV*s)(3.00 x 10^8 m/s) / (12.75 eV)
= 9.74 x 10^-8 m

I am just unsure of my work and would really like a second opinion. Thank you!
 
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To simplify the calculations in d, you can equate hc=1240 eV nm. Rest everything seems ok to me. (I haven't checked your calculations.)
 
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