## Rydberg's Equation [ENDORSED]

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Protonated-Pam1D
Posts: 16
Joined: Tue Mar 07, 2017 3:03 am

### Rydberg's Equation

For Problem 1.13 why is it that the equation used to solve for the wavelength is 1/lambda = R ( 1/n^2 - 1/n^2)? Why isn't it E = - R (1/n^2 + 1/n^2) and then use E = h x c/ lambda to solve for the wavelength?

alexis castro 1B
Posts: 30
Joined: Wed Nov 16, 2016 3:02 am

### Re: Rydberg's Equation

What Lavelle has taught us in class is the empirical version of the Rydberg equation. You can still use the equation lavelle showed us in class to solve the problem but it would just take a few more steps than how the solutions manual showed it with just plugging it all into the original version of the rydberg equation.

Diego Zavala 2I
Posts: 65
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

### Re: Rydberg's Equation  [ENDORSED]

Because E=hvand E=hR[z], the first equation can be rewritten as v=R[z]. Also because (Lambda)x(frequency)=c, Bohr's equation can be written as c/(Lambda)=R[z].

z=(1/n(1)^2)-(1/n(2)^2).