alleryEnergy of the photon emitted in a hydrogen transition from n=4 to n=2:
A13.6 × (1/4 - 1/16) = 2.55 eV
B-13.6 / (16 - 4)
C13.6 / 4
D13.6 × (1/4 - 1/16)
Answer & Solution
Correct answer: A. 13.6 × (1/4 - 1/16) = 2.55 eV
We have ΔE = E_2 - E_4.
That is ΔE = -13.6/4 - (-13.6/16).
That is ΔE = -13.6 × (1/4 - 1/16).
That is ΔE = -13.6 × (4-1)/16.
So ΔE = -2.55 eV.
Photon energy = |ΔE| = 2.55 eV.
Wavelength about 486 nm (Hβ line, blue-green visible).
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