## Atomic Spectrum Module #27

$c=\lambda v$

melodyzaki2E
Posts: 61
Joined: Fri Sep 28, 2018 12:18 am

### Atomic Spectrum Module #27

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?

How is the first question different from what's given in the beginning of the problem?

ThomasLai1D
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: Atomic Spectrum Module #27

I believe what's given is the number of wavelengths per meter (1,650,763.73/m). So if you take the inverse of the number, you should be able to find the wavelength (1/1,650,763.73*m^-1 = 6.05x10^-7 m)