Origin of electron in Paschen series (hydrogen atom spectrum)?

[rainingspiders]

I keep getting n = 3 for the origin, but the answer is n = 6... what am I doing wrong?

 

Question: Given that the wave number of a line in the Paschen series of hydrogen is 9140 cm^-1, what is the origin of the electron (what energy state?)

 

My method:

 

wave number = 1/(wavelength) = 9140 cm^-1 = 91.4 m^-1

lowest state for Paschen series is n = 3, so n-final = 3

 

Since this is emission...

 

E = (2.1799 x 10^-18)(1/(n-final)^2 - 1/(n-initial)^2)

 

(plank's constant) * (speed of light) * (1/(wavelength)) = (2.1799 x 10^-18)(1/(3)^2 - 1/(initial state)^2)

 

Solving for n-initial...

n-initial =3.

 

I know this is wrong, but I don't know what I did wrong. Help?

 

[Sensei]

Calculation of energy of photons emitted/absorbed by Hydrogen is:

 

[math]E=\frac{13.6 eV}{n^2}-\frac{13.6 eV}{m^2}[/math]

 

[math]13.6 eV * 1.6021766*10^{-19}=2.17896*10^{-18} J[/math]

 

When n=3,

and m>=4

then:

 

[math]E=\frac{13.6 eV}{3^2}-\frac{13.6 eV}{4^2}=0.66111 eV[/math]

 

[math]E=\frac{13.6 eV}{3^2}-\frac{13.6 eV}{5^2}=0.96711 eV[/math]

 

[math]E=\frac{13.6 eV}{3^2}-\frac{13.6 eV}{6^2}=1.13333 eV[/math]

 

To convert eV to J there is needed to multiply by e.

 

[math]E=\frac{hc}{\lambda}[/math]

 

Convert energy to wavelength:

[math]\lambda=\frac{hc}{E}[/math]

Edited November 12, 2014 by Sensei

[rainingspiders]

wait, 2.1786 x 10^-18 J?

 

I thought the numerator should be 2.1799 x 10^-18 J?

[swansont]

 

wave number = 1/(wavelength) = 9140 cm^-1 = 91.4 m^-1

 

 

…

 

I know this is wrong, but I don't know what I did wrong. Help?

 

 

Check your math here. 9140 cm = 91.4 m, but 9140 cm^-1 ≠ 91.4 m^-1

[rainingspiders]

@swansont:

 

Oh, I see what I did wrong now. Thanks!