Emission & Energy levels of Hydrogen Problem

AI Thread Summary
The discussion focuses on determining the energy levels involved in the emission of a 388-nm photon from hydrogen gas. The calculated energy of the photon is approximately 3.206 eV, leading to the conclusion that the transition occurs from n=8 to n=2. The Rydberg formula is highlighted as essential for identifying the initial and final energy levels in hydrogen transitions. Participants emphasize the importance of understanding the relationship between wavelength and energy levels in hydrogen. The key takeaway is that the transition responsible for the photon emission starts from n=8.
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Homework Statement



What values of n are involved in the transition that gives the rise to the emission of a 388-nm photon from hydrogen gas?


Homework Equations



n=?
wavelength = 388nm=3.88x10^-7 m
R=1.097x10^7 m^-1

E= hc/ wavelength


The Attempt at a Solution



E= hc/ wavelength=(6.63x10^-34)(3.00x10^8)/ (3.88x10^-7) = 5.13x10^-19 J = 3.206 eV

En = -13.6eV / n^2
n = - square root of (13.6/En) = -square root of (13.6/3.206) = - square root of 4.24 = -2.06

n = 2 <- 1st excited state


the answer is n=8 to n= 2.

How would I know that it began from n=8 ?? Maybe there's an equation that I missed..
 
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The formula you need to know for this is the rydberg formula.

\frac{1}{\lambda}=R_H \left(\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}\right) where, n_1&lt;n_2
 
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