Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 6 to the n = 2 energy level. h = 6.63 x 10^-34 J s, RH = 2.18 x 10^-18 J, E = hc/λ = -RH (1/nf^2 – 1/ni^2)
Added by Noah R.
Step 1
Substituting the given values: E = -2.18 x 10^-18 J (1/2^2 – 1/6^2) = -2.18 x 10^-18 J (1/4 – 1/36) = -2.18 x 10^-18 J (0.25 - 0.0278) = -2.18 x 10^-18 J * 0.2222 = -4.84 x 10^-19 J The energy difference is negative because energy is being emitted as the Show more…
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