Question

Calculate the wavelengths of the first five members of the lyman series of spectral lines

Answer 1

Answer:

  λ₂ = 1,219 10⁻⁷ m , λ₃ = 1.028 10⁻⁷ m ,   λ₄ = 0.9741 10⁻⁷ m , λ₅ = 0.9510 10⁻⁷ m and  λ₆ = 0.9395 10⁻⁷ m

Explanation:

To calculate the lines of the hydrogen liman series, we can use the Bohr atom equation

           En = -13.606 / n²       [eV]

n       En

1       -13,606

2       -13.606 / 4 =    -3.4015

3       -13.606 / 9 =    -1.5118

4       -13.606 / 16 =  -0.8504

5       -13.606 / 25 = -0.5442

6       -13.606 / 36 = -0.3779

The lyma series are transitions where the state is fundamental (E1), let's calculate the first five transitions

State

initial final energy

6           1      -0.3779 - (- 13.606) =  13.23 eV

5           1      -0.5442 - (- 13.606) =  13.06 eV

4           1      -0.8504- (-13.606) =   12.76 eV

3            1      -1.5118 - (- 13.606) =   12.09 eV

2            1      -3.4015 - (- 13.606) = 10.20 eV

Let's use the relationship between the speed of light and the wavelength and the frequency

      c = λ  f

      f = c / λ  

Planck's relationship for energy

     E = h f

     E = h c / λ

    λ = hc / E

We calculate for each energy

E = 10.20 eV

      λ  = 6.63 10⁻³⁴ 3 10⁸ / (10.20 1.6 10⁻¹⁹)

      λ  = 12.43 10⁻⁷ / 10.20

      λ₂ = 1,219 10⁻⁷ m

E = 12.09 eV

     λ₃ = 12.43 10⁻⁷ / 12.09

     λ₃ = 1.028 10⁻⁷ m

E = 12.76 eV

      λ₄ = 12.43 10⁻⁷ /12.76

      λ₄ = 0.9741 10⁻⁷ m

E = 13.06 ev

      λ₅=  12.43 10⁻⁷ /13.06

       λ₅ = 0.9510 10⁻⁷ m

E = 13.23 eV

      λ₆ = 12.43 10⁻⁷ / 13.23

      λ₆ = 0.9395 10⁻⁷ m