Question

An electron in the hydrogen atom jumps from an orbit in which the energy is 2.86 eV higher than the energy of the final lower-energy orbit. (a) What is the frequency of the photon emitted in this transition? (h=4.14×10^-15 eV·s.) (b) What is the wavelength of the emitted photon?

Name: an electron in the hydrogen atom jumps from an orbit in which the energy is 286 ev higher than the energy of the final lower energy orbit a what is the frequency of the photon emitted in thi 93221
Uploaded: 2022-05-25T07:08:07-08:00
Duration: 1 min 51 s
Channel: Suman K
Description: an electron in the hydrogen atom jumps from an orbit in which the energy is 286 ev higher than the energy of the final lower energy orbit a what is the frequency of the photon emitted in thi 93221

          An electron in the hydrogen atom jumps from an orbit in which the energy is 2.86 eV higher than the energy of the final lower-energy orbit. (a) What is the frequency of the photon emitted in this transition?
(h=4.14×10^-15 eV·s.) (b) What is the wavelength of the emitted photon?

Added by Trinidad A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Suman K

05/25/2022

Step 1

Given: Change in energy (ΔE) = 2.86 eV, Planck's constant (h) = 4.14 × 10^-15 eV·s Frequency (f) = ΔE / h Frequency (f) = 2.86 eV / (4.14 × 10^-15 eV·s) Frequency (f) = 6.906 × 10^14 Hz Show more…

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An electron in the hydrogen atom jumps from an orbit in which the energy is 2.86 eV higher than the energy of the final lower-energy orbit. (a) What is the frequency of the photon emitted in this transition? (h=4.14×10^-15 eV·s.) (b) What is the wavelength of the emitted photon?