From the figure below, we can see that the Lyman series corresponds to transitio

Question

From the figure below, we can see that the Lyman series corresponds to transitions ending at the ground-state energy, Ef = E1 = -13.6eV. Because the photon wavelength varies inversely with energy, the transition that has the longest wavelength is the transition that has the lowest energy, which is that from the first excited state n = 2 to the ground state n = 1. The longest wavelength of the Lyman series for the hydrogen atom was calculated to be 122 nm in Example 36-2 of the textbook.

Find the wavelengths for the following transitions.

From the figure below, we can see that the Lyman series corresponds to transitions ending at the ground-state energy, Ef = E1 = -13.6eV. Because the photon wavelength lambda varies inversely with energy, the transition that has the longest wavelength is the transition that has the lowest energy, which is that from the first excited state n = 2 to the ground state n = 1. The longest wavelength of the Lyman series for the hydrogen atom was calculated to be 122 nm in Example 36-2 of the textbook. Find the wavelengths for the following transitions. (a) n1 = 5 to n2 = 1 nm (b) n1 = 7 to n2 = 1 nm

Explanation / Answer

( a ) 

E = k / n^2 ; 

-13.6 = k / 1^2 ; 

k = -13 .6 

E5 =- 13.6 / 5^2 = - 0.544eV ; 

E1 = - 13.6 eV ; 

change in energy = -13.6 -  - 0.544 = 13.056 eV ; 

13.056 eV = hf ; 

13.056 * 1.6 e -19  =  h c /  ; 

= [hc ] / [ 13.056 * 1.6 e -19 ]

= 9.50925 e -8 m   <----------ans 

= 95.0925 e -9 m 

= 95.0925 nm   <--------ans

( b ) 

E = -13.6 / 1^2  -  - 13.6 / 7^2  = -13.322 eV ; 

13.322 eV = hc /  ; 

= [ hc ]  /  [ 13.322 * 1.6 e -19 ]

= 9.31938 e -8    <--------------ans

= 93.1938 nm   <-----------ans

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