## energy of light

$c=\lambda v$

505166714
Posts: 50
Joined: Fri Sep 28, 2018 12:17 am

### energy of light

When excited electrons fall back from the excited state to ground state, what determines the energy besides frequency and wavelength? Will the number of electrons falling back affects the amount of energy given off?

To see a specific question: go to exercise 1.25 in the 6th edition book.

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

### Re: energy of light

for (a), you would use the equations c=(wavelength)*(freq) and E=h*(freq) to derive the equation E=h*c/(wavelength). Because you are given h, c, and wavelength, you can calculate the energy given off by an excited sodium atom. Using this energy calculation, you can answer parts (b) and (c) by converting the given amounts into the number of atoms (using Avogadro's number). The energy amount would then be calculated using the number of atoms multiplied by the original energy per atom found in part (a)

505166714
Posts: 50
Joined: Fri Sep 28, 2018 12:17 am

### Re: energy of light

If the number of particles determines the amount of energy given off, then the assumptions in the photoelectric effect example would be wrong, since we can manipulate the number of photons to excite electrons no matter what the frequency is. How is this possible?