Atomic Spectra Post Assessment Question

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Taylor_Yamane_2L
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Atomic Spectra Post Assessment Question

Postby Taylor_Yamane_2L » Fri Sep 30, 2016 10:13 pm

A question on the atomic spectra module post assessment asks the following question and I'm am not exactly sure how to figure it out. Any hints would be much appreciated! (:

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?

A. 1.651 x 106 m (1651 km), Ultraviolet light,1.204 x 10-31 J

B. 6.058 x 10-7 m (605.8 nm), Infrared light,2.491 x 1039 J

C. 6.058 x 10-7 m (605.8 nm), Visible light,4.14 x 10-40 J

D. 6.058 x 10-7 m (605.8 nm), Visible light,3.281 x 10-19 J

Edward_Lee_3C
Posts: 36
Joined: Wed Sep 21, 2016 2:58 pm

Re: Atomic Spectra Post Assessment Question

Postby Edward_Lee_3C » Sat Oct 01, 2016 5:21 pm

I'm not too sure as well but since 1,650,763.73 wavelengths equals 1 meter. You divide 1 meter by 1,650,763.73 to get 6.06*10^-7. If you check the wavelength chart, it is in the visible light region and to find the energy you can use E=(hc)/wavelength.

Chin_Alyssa_3I
Posts: 34
Joined: Fri Jul 22, 2016 3:00 am

Re: Atomic Spectra Post Assessment Question

Postby Chin_Alyssa_3I » Mon Oct 03, 2016 12:01 am

Where did you get 1 meter from?

Chem_Mod
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Re: Atomic Spectra Post Assessment Question

Postby Chem_Mod » Mon Oct 03, 2016 11:48 am

From the definition of a meter provided in the question.


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