If an electron in a hydrogen atom relaxes to the second energy level emitting a photon of wavelength 410.2 nm what higher energy level did it start at?

Answer 1

7

Answer 2

There are actually two ways to do this. I'll explain how to use Rydberg's formula, though, because it's significantly simpler to use.

Rydberg's Formula states: 1/(wavelength)=R (1/n1^2-1/n2^2)

R= 1.097x10^7m^-1 (this is Rydberg's constant so you can look this up)

n1=nfinal (or n=2 in this problem)

n2=ninitial (what we are looking to find)

wavelength=lambda=3.97x10^-7m

Set up the problem:

1/3.97x10^-7m=(1.097x10^7m^-1)*(1/2^2-1/n^2)

.2296=1/4-1/n^2

1=.02n^2

n^2=50

n=~7

Alternatively you could have used Bohr's model but that would take two formulas and a lot more work for the same answer (using Bohr's model is simply working backwards from what it is typically used for).

Answer 3

The hydrogen atom electron likely began at the third energy level before transitioning to the second energy level. This is because the transition from the third to the second energy level corresponds to a photon with a wavelength of 410.2 nm in the Balmer series.

Answer 4

6