Rydberg's Equation  [ENDORSED]

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Protonated-Pam1D
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Rydberg's Equation

Postby Protonated-Pam1D » Thu Jul 06, 2017 12:17 am

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

Postby alexis castro 1B » Thu Jul 06, 2017 10:04 pm

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
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Re: Rydberg's Equation  [ENDORSED]

Postby Diego Zavala 2I » Sat Oct 07, 2017 10:48 pm

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).


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