## Atomic Spectra Post-Module #28

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

tiffanyteguh1C
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Joined: Fri Sep 29, 2017 7:04 am

### Atomic Spectra Post-Module #28

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 do we find the new wavelength?

Joyce Lee 1C
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Joined: Fri Sep 29, 2017 7:03 am
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### Re: Atomic Spectra Post-Module #28

To find the wavelength with the given information, divide 1m by 1,650,763.73 wavelengths.
The length of a single wave is 6.058 x 10^-7 m or 605.8 nm
This wavelength corresponds to visible light, which has the range of about 400-700 nm.
To find the energy of one photon, use the formula E = hc/lambda

E = [(6.626 x 10^-34 Js)(3.00 x 10^8 m/s)]/(6.058 x 10^-7 m) = 3.281 x 10^-19 J