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Emission & Energy levels of Hydrogen Problem

1. 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?


2. Homework Equations

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

E= hc/ wavelength


3. 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..
 

Kurdt

Staff Emeritus
Science Advisor
Gold Member
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The formula you need to know for this is the rydberg formula.

[tex]\frac{1}{\lambda}=R_H \left(\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}\right)[/tex] where, [tex]n_1<n_2[/tex]
 
Last edited:

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