00:01
All right guys the problem uh the 138 chapter 7 so bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589 and 589 .6 nanometers where the frequency in 8 and energy of a photon if at each uh at each of these wave lanes and what's the energy in kilojou's per model so we have the frequency a we're going to use mu is equal to lambda over c over wave sorry mu is going to be equal to c over lambda with mu equal frequency c is speed of light and lambda's wavelength.
00:34
So we have these two set up.
00:35
We plug our numbers in and we're going to get 5 .093 times 10 to the 14th inverse seconds and we got 5 .088 times 10 to the 14 inverse seconds.
01:07
So then we got how to find the energy.
01:11
So e1 is going to equal h mu1, which is going to be equal to 6 .63 times 10 to the negative 34 inverse seconds.
01:28
Sorry, negative 34, not inverse seconds.
01:31
Jules seconds times 5 .093 times 10 to the.
01:42
14 inverse seconds and then e2 that's going to equal 6 .63 times 10 to the negative 34 joules, joules seconds i should say over 5 .08 times 10 to the 14 inverse seconds.
02:24
This is going to be equal to 3 .37 times 10 to the negative 19 joules...