Question

An electron in the hydrogen atom jumps from an orbit in which the energy is 3.1 eV higher than the energy of the final lower-energy orbit. Take Planck's constant h as 4.14 × 10^-15 eV·s and the velocity of light as 3 × 10^8 m/s. What is the wavelength of the emitted photon? (Round the final answer to four decimal places.)

Name: an electron in the hydrogen atom jumps from an orbit in which the energy is 31 ev higher than the energy of the final lower energy orbit take plancks constant h as 414 1015 evs and the veloc 99727
Uploaded: 2022-05-17T13:40:12-08:00
Duration: 2 min 33 s
Channel: Adi S
Description: an electron in the hydrogen atom jumps from an orbit in which the energy is 31 ev higher than the energy of the final lower energy orbit take plancks constant h as 414 1015 evs and the veloc 99727

          An electron in the hydrogen atom jumps from an orbit in which the energy is 3.1 eV higher than the energy of the final lower-energy orbit. Take Planck's constant h as 4.14 × 10^-15 eV·s and the velocity of light as 3 × 10^8 m/s.
What is the wavelength of the emitted photon? (Round the final answer to four decimal places.)

Added by Jose Angel T.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

05/17/2022

Step 1

1 eV. We can convert this energy into joules using the conversion factor 1 eV = 1.602 x 10^-19 J. Show more…

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An electron in the hydrogen atom jumps from an orbit in which the energy is 3.1 eV higher than the energy of the final lower-energy orbit. Take Planck's constant h as 4.14 × 10^-15 eV·s and the velocity of light as 3 × 10^8 m/s. What is the wavelength of the emitted photon? (Round the final answer to four decimal places.)