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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 :

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

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