Question

For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equation

1/?= 2p2 mk2 e4/ h3c (Z2) (1/n2f - 1/n2i)

Using this equation, show that the ranges of wavelengths in these two series overlap.
Shortest Wavelength (m) Longest Wavelength (m)

Paschen Series

Answer 1

Answer:

the highest value of the Paschen series is 18805 μm which is greater than the shortest value of the Brackett series (14617 μm).

Explanation:

Hydrogen atom transitions are described by the Bohr model

        $E_{n}$ = -13.606 / n²

Where n is an integer

Transitions occur between two states with different quantum numbers n

         ΔE = -13.606 (1 / $n_{f}$² - 1 / $n_{i}$²)

Where $n_{f}$ > $n_{i}$

For the wavelength we use the Planck equation

         E = h f

         c = λ f

          E = h c / λ      

         λ = h c / E

         λ = 6.63 10⁻³⁴ 3 10⁸ / E

         λ = 19.89 10⁻²⁶ / E

Let's reduce to eV

         λ = 19.89 10⁻²⁶ / E[J]  (Ev / 1.6 10⁻¹⁹ J)

         Lam = 12430 / E [eV]

 

Let's calculate the energy for transitions

Paschen Series

$n_{f}$     $n_{i}$     ΔE (eV)       λ  (um)

4       3       0.661          18805

5        3       0.9657        12871

10       3       1.3757          9035

100     3      1.5104           8229

∞       3      1.5117            8222

Brackett Series

$n_{f}$    $n_{i}$        ΔE (eV)        λ  (um)

5      4         0.3061        40607

6      4         0.4724        26312

10    4          0.7143         17401

100  4         0.8490        14640

∞    4         0.8504        14617

When we analyze the wavelength values ​​we see that the highest value of the Paschen series is 18805 μm which is greater than the shortest value of the Brackett series (14617 μm).