Question

Excited H atoms have many emission lines. One series of lines, called the brackett series, occurs in the infrared region. It results when an electron changes from higher energy levels to a level with n=4. Calculate the wavelength and frequency of the lowest energy line of this series.

Answer 1

Lowest energy occurs when electron jumps from n=5 to n=4 in Brackett series.

From Rydburg's Equation , 1/ λ = R[(1/n_i²) – (1/n_f²)]
   Where R = Rydburg's constant = 10.96 x10⁶ m^-1
               λ = wavelength = ?
               n_i = 4
               n_f = 5
Plug the values we get

1/ λ = R[(1/n_i²) – (1/n_f²)]

       = ( 10.96 x10⁶ m^-1 )x[(1/4²) – (1/5²)]

       = 246600

    λ = 4.05x10^-6 m

So the wavelength is 4.05x10^-6 m

We knpow that frequency , = c/λ

Where

c = speed of light = 3x10⁸ m/s

Plug the values we get

= c/λ

   = (3x10⁸ ) / (4.05x10^-6 )

   = 7.398x10¹³ s^-1

Therefore the frequency is 7.398x10¹³ s^-1