## post assessment #28

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

Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### post assessment #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?

I don't understand why you have to divide the 1,650,763.73 wavelengths into 1.
Also when I solved for the energy of one photon, I got 3.28 x 10^-42 and the answer is 3.28 x 10 ^ -19 which I don't understand why.

305113590
Posts: 66
Joined: Fri Sep 28, 2018 12:28 am

### Re: post assessment #28

1 meter has 1,650,763.73 wavelengths. If you divide it, you will get 6. 0578x10^-7. This number means that in one wavelength (trough to trough) it is equivalent to 605.8nm. I do not think that energy was a variable to be considered in the problem.

For part two of your question here is the math:

c=λv
3x10^8=(6.0578x10^-7)(v)

v equals 4.9523x10^14

Then you plug it in the E=hv where you solve for E.

E=(6.626x1-^-34)(4.9523x10^14)= 3.2814x10^-19