Atomic Spectra Module

[Mrudula Akkinepally]

Hi everyone!

I came across this question while taking the assessments associated with the modules on Dr. Lavelle's website and I'm not sure how to solve this. I am pasting the question below!

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?

[Stuti Pradhan 2J]

Since you are told that 1 meter is 1,650,763.73 wavelengths of radiation emitted by krypton-86, it is basically a 1 meter to 1,650,763.73 wavelengths ratio. To find 1 wavelength, just divide 1 meter by 1,650,763.73.

If you set it up as 1m=1,650,763.73 wavelengths, then your goal is to find 1 wavelength, so divide both sides by 1,650,763.73 to get your answer.

Once you find the wavelength, you can figure out what part of the electromagnetic spectrum it corresponds to. You can also use the equation to find the energy of one photon.

Hope this helps!

[Olivia Monroy 1A]

To put it simply divide 1 wave by 1,650,763.73 m to get 6.0578x10^-7 m per wavelength then with that you can find the energy per photon using E=hc/wavelength to get an answer of 3.281x10^-19 J, matching option D.

[Kimiya Aframian IB]

Hi everyone!

I came across this question while taking the assessments associated with the modules on Dr. Lavelle's website and I'm not sure how to solve this. I am pasting the question below!

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?

Hi! For me, the trouble in this problems was not exactly the calculations but instead seeing what is was asking us to do. I think that it is pretty much saying that in one meter, there are 1650763.73 waves, so we need to divide the two in order to solve for the wavelength, which we can then use to solve for energy.

Hope this helps!