## Thursday, August 21, 2014

### 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 1st, 2nd and 3rd Balmer transitions will be, respectively:

1st)  E (n=5 to n = 2 ) =  - 13.6 ( 1/ 5 2   -1/ 2 2 )

2nd) E (n=4 to n = 2 ) =  - 13.6 ( 1/ 4 2   -1/ 2 2 )

3rd)  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