Question

An electron in the hydrogen atom jumps from an orbit in which the energy is 3.36 eV higher than the energy of the final lowering energy orbit. Take Planck's constant h as 4.14 * 10^-15 eVs and the velocity of light as 310^8 m/s. Calculate the wavelength of the emitted photon. Answer: The wavelength of the emitted photon is _______ nm.

Name: an electron in the hydrogen atom jumps from an orbit in which the energy is 336 ev higher than the energy of the final lowering energy orbit take plancks constant h as 414 10 15 evs and the 25458
Uploaded: 2022-06-22T22:38:08-08:00
Duration: 1 min 16 s
Channel: Penny Riley
Description: an electron in the hydrogen atom jumps from an orbit in which the energy is 336 ev higher than the energy of the final lowering energy orbit take plancks constant h as 414 10 15 evs and the 25458

          An electron in the hydrogen atom jumps from an orbit in which the energy is 3.36 eV higher than the energy of the final lowering 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.
Calculate the wavelength of the emitted photon.
Answer: The wavelength of the emitted photon is _______ nm.

Added by Christian S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Penny Riley

06/22/2022

Step 1

36 eV Planck's constant (h) = 4.14 * 10^-15 eV*s Speed of light (c) = 3 * 10^8 m/s Show more…

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An electron in the hydrogen atom jumps from an orbit in which the energy is 3.36 eV higher than the energy of the final lowering 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. Calculate the wavelength of the emitted photon. Answer: The wavelength of the emitted photon is _______ nm.