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3) A photon is emitted as the electron in a hydrogen
atom drops from the n = 5 energy level directly to
the n = 3 energy level. What is the energy of the
emitted photon?
(1) 0.85 eV
(2) 0.97 eV
(3) 1.51 eV
(4) 2.05 eV

Sagot :

pstnonsonjoku

From the information provided and the calculations, the energy of the drop is 1.51 eV

What is the Rydberg formula?

The Rydberg formula is used to obtain the energy of a photon or the wavelength of a photon.

Thus;

1/λ = 1.097 * 10^7 (1/3^2 - 1/5^2)

1/λ = 1.097 * 10^7(0.11 - 0.04)

λ = 1.3 * 10^-6 m

Since;

E = hc/λ

E = 6.6 * 10^-34 * 3 * 10^8/1.3 * 10^-6

E = 1.51 * 10^-19 J or 1.51 eV

Learn more about photon:https://brainly.com/question/20912241

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