## Atomic Spectra Post-Module, Question 17 [ENDORSED]

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

melissagonzalez1c
Posts: 7
Joined: Wed Sep 21, 2016 3:00 pm

### Atomic Spectra Post-Module, Question 17

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?

I don't really understand what the question means when it says this, what does the meter have to do with the wavelength emitted by krypton-86? I may just be reading the question wrong, if anyone could clear this up it would be greatly appreciated!

Christopher Reed 1H
Posts: 33
Joined: Wed Sep 21, 2016 2:57 pm
Been upvoted: 1 time

### Re: Atomic Spectra Post-Module, Question 17  [ENDORSED]

Think of the question asking you to find the length of each individual "part" that makes up one meter, given the total number of "parts." You would do this by dividing (partitioning) one meter by the total number of parts given.
Having said this, replace the word part with "wavelength" and hopefully it will make more sense. (:

$\frac{1m}{1,650,763.73} = 6.058 x10^{-7}m$

The above calculation is for one wavelength emitted by krypton-86!

melissagonzalez1c
Posts: 7
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Atomic Spectra Post-Module, Question 17

Christopher Reed 4I wrote:Think of the question asking you to find the length of each individual "part" that makes up one meter, given the total number of "parts." You would do this by dividing (partitioning) one meter by the total number of parts given.
Having said this, replace the word part with "wavelength" and hopefully it will make more sense. (:

$\frac{1m}{1,650,763.73} = 6.058 x10^{-7}m$

The above calculation is for one wavelength emitted by krypton-86!

Thank you so much!! It makes a lot more sense now :~)

### Who is online

Users browsing this forum: No registered users and 1 guest