**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

**for Balmer lines:***differences between energy levels*
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

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

(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 H__

__g__

__):__

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

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

(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
## No comments:

Post a Comment