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

User CrabMan
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1 Answer

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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

Step-by-step 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

User Sfink
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