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
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 )
(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
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 )
(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
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