alleryWhen a hydrogen atom makes a transition from n = 2 to n = 1, the energy of the emitted photon is approximately:
A10.2 eV
B3.4 eV
C1.51 eV
D13.6 eV
Answer & Solution
Correct answer: A. 10.2 eV
We have ΔE = E_2 − E_1.
That is ΔE = (−3.4) − (−13.6).
So ΔE = 10.2 eV.
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