## Rydberg Formula for Atomic Hydrogen [ENDORSED]

$c=\lambda v$

Cindy Nguyen 1L
Posts: 48
Joined: Fri Apr 06, 2018 11:04 am

### Rydberg Formula for Atomic Hydrogen

On the front page of Test 1, the Rydberg Formula is $E_{n} = -hRhn^{2}$. In lecture, it is $E_{n}=\frac{-hR}{n^{2}}$. I was wondering how the formula on Test 1 is equivalent to the one given in lecture.

LilianKhosravi_1H
Posts: 32
Joined: Fri Apr 06, 2018 11:03 am

### Re: Rydberg Formula for Atomic Hydrogen

The one on my cover of test 1 is -(hR/n^2), which is the same as the one given in lecture.

Sarah Brecher 1I
Posts: 31
Joined: Fri Apr 06, 2018 11:02 am

### Re: Rydberg Formula for Atomic Hydrogen

While the test will give you the E(n) = -(hR)/n^2, a faster equation that I was given in discussion that is very helpful is:

v (frequency) = R [(1/nf^2) - (1/ni^2)]

Endri Dis 1J
Posts: 33
Joined: Fri Apr 06, 2018 11:02 am

### Re: Rydberg Formula for Atomic Hydrogen

There are actually many forms of Rydberg formula given but the one that I use is

DeltaE = -R*((1/nf^2)-(1/ni^2))*Z^2

Where
DeltaE= diff in energy between the final and initial energy level(J)
R= Rydberg"s compound(2.178X10^-18J)
nf= final energy level
ni= initial energy level
Z=atomic# (# of protons)

Also, Rydberg's constant can be written as

2.178X10^-18J
3.29X10^15s^-1
1.097X10^7m^-1

Erin Li 1K
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am

### Re: Rydberg Formula for Atomic Hydrogen  [ENDORSED]

it's E(n) = -(hR)/nfinal^2 -(hR)/ninitial^2
and for frequency it is = R [(1/nfinal^2) - (1/ninitial^2)]