Question

2. Calculate the energy of an electron in the n=2 energy level of hydrogen. Calculate the energy of an electron in the n=3 energy level. What is the difference in energy of these two levels? If a photon of -2,18 X/ 08 2427 light had this energy, what would its wavelength be?

Answer 1

The energy level of an electron in a hydrogen atom = E_n

E_n = -13.6/n² eV (n = orbit number)

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The energy of an electron in the n = 2 energy level of hydrogen = E₂

E₂ = -13.6/2² eV = -13.6/4 eV = -3.4 eV

(E₂ = -3.4 eV)

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The energy of an electron in the n = 3 energy level of hydrogen = E₃

E₃ = -13.6/3² eV = -13.6/9 eV = -1.51 eV

(E₃ = -1.51 eV)

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Difference in energy of these two levels = ∆E

∆E = E3 - E2

∆E = -1.51 -(-3.4) eV = 1.89 eV

( ∆E = 1.89 eV )

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1 eV = 1.602×10^-19 J

So, 1.89 eV = 1.89×1.602×10^-19 = 3.028×10^-19 J

∆E = 3.028×10^-19 J

Let's say the wavelength of the photon is λ.

∆E = hc/λ (h = planck's constant, c = speed of light)

3.028×10^-19 J = (6.626×10^-34 J s)(3×10⁸ m/s)/λ

λ = (6.626×3/3.028)×10^-34+8+19 m

λ = 6.565×10^-7 m

λ = 656.5 nm

( Wavelength of a photon is 656.5 nm )