Basic Atomic Physics: Insight Problem Solution

We now look at the solution of the insight problem at the end of the Basic Atomic Physics blog post:

Solution:

The energies for the 1^st, 2^nd and 3^rd Balmer transitions will be, respectively:

 

1^st)  E _{(n=5 to n = 2 )} =  - 13.6 ( 1/ 5 ²   -1/ 2 ² )

 

2^nd) E _{(n=4 to n = 2 )} =  - 13.6 ( 1/ 4 ²   -1/ 2 ² )

 

3^rd)  E _{(n=3 to n = 2 )} =  - 13.6 ( 1/ 3 ²   -  1/ 2 ² )

 

Take differences between energy levels for Balmer lines:

 

Balmer a line (called H- alpha):

 

E3 – E2 =  - 13.6 eV ( 1/ 3 ²   -  1/ 2 ² )   = -

 

 13.6 eV( 1/9 – ¼) = -13.6 eV (-5/ 36) = 1.88 eV

 

Now, 1 eV =  1.6 x 10^-19 J  so:

 

So:  E3 – E2 =    1.88 eV  (1.6 x 10^-19 J  /eV) = 3.02 x 10^-19 J 

 

From this, the wavelength of the photon emitted can be found. Since E = hf = h (c/ l):

l =   hc/ (E3 – E2) 

 

l =    (6.626069 x 10^{- 34} J-s)(3 x 10 ⁸ m/s)/ (3.02 x 10^-19 J )   

 

l =    6.56 x  10^{- 7} m

 

Balmer b line (called H _b):

E4 – E2 =  - 13.6 eV ( 1/ 4 ²   -  1/ 2 ² )   

 

= - 13.6 eV( 1/16 – ¼) =   -13.6 eV ( -3/16) =    2.55 eV

 

 1 eV =  1.6 x 10^-19 J  so:

 

So:  E4 – E2 =    

2.55 eV (1.6 x 10^-19 J  /eV) = 4.08 x 10^-19 J 

 

As before, the wavelength of the photon emitted is:

l =   hc/ (E4 – E2)  = 

 (6.626069 x 10^{- 34} J-s)(3 x 10 ⁸ m/s)/ (4.08 x 10^-19 J )

 

l =    4.47 x  10^{- 7} m

 

Balmer g  line (called Hg ):

 

E5 – E2 =  - 13.6 eV ( 1/ 5 ²   -  1/ 2 ² )  

 

= - 13.6 eV( 1/25 – ¼) = -13.6 eV ( -21/100) =    3.4 eV

 

 1 eV =  1.6 x 10^-19 J  so:

 

So:  E5 – E2 =   

 3.4  eV (1.6 x 10^-19 J  /eV) = 5.44 x 10^-19 J 

 

As before, the wavelength of the photon emitted is inversely proportional to the difference between energy levels:

 

l =   hc/ (E5 – E2)  = 

(6.626069 x 10^{- 34} J-s)(3 x 10 ⁸ m/s)/ (5.44 x 10^-19 J )

 

l =    3.63 x  10^{- 7} m