Atomic spectra pre assessment

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Ekin 1C
Posts: 21
Joined: Fri Sep 29, 2017 7:04 am

Atomic spectra pre assessment

Postby Ekin 1C » Fri Oct 13, 2017 9:33 pm

Does anyone understand number 27 of the pre-assessment for atomic spectra
The question is, "The meter was defined in 1963 as 1,650,763.73 wavelengths of radiation emitted by krypton-86 (it has since been redefined). What is the wavelength of this krypton-86 radiation? To what region of the electromagnetic spectrum does this wavelength correspond (i.e. infrared, ultraviolet, x-ray, etc.)? What energy does one photon of this radiation have?"

miznaakbar
Posts: 55
Joined: Thu Jul 13, 2017 3:00 am
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Re: Atomic spectra pre assessment

Postby miznaakbar » Fri Oct 13, 2017 11:33 pm

To solve for the wavelength of the Krypton, you need to divide 1,650,763.73 wavelengths into 1 meter (1/1,650,763.73)- this will give you the length in meters of each wavelength, which should be 6.06x10^-7m (606nm, which is in the visible light region). Next to find the energy of one photon, you need to use E = hv, and since v=(c/wavelength), you can use E=h(c/wavelength). Plugging in plank's constant and the speed of light while using the 6.06x10^-7 m will give us 3.28 x 10^-19 Joules as the answer.


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