Question

You have an atom of Chucktanium, for which the energy levels are:

n = 1 (ground state) EPE = -100.0 eV
n = 2 (first excited state) EPE = -82 eV
You observe a photon emitted from the electron transition n = 2 down to n = 1. Calculate the frequency of that outgoing photon, as a multiple of 1015 Hz. E.g., if your answer is 2.0379 × 1015 Hz, then type in 2.04 for your answer.

For simplicity: Work in electron-Volts, using h = 4.136×10-15eV sec.

Answer 1

U know
E=h*f

Answer 2

But E=EPE2-EPE1

Answer 3

So
EPE2-EPE1=h*f
-82-(-100)=4.136*10^-15*f
f=?