Question

(b). Determine the (i) longest and (ii) shortest wavelength lines (in nanometers) in the Paschen series of the hydrogen spectrum. In which region of the spectrum is the shortest- wavelength line? (c) Using the Balmer-Rydberg equation, calculate the value of n corresponding to the violet emission line (wavelength = 434.0 nm) in the Balmer series of the hydrogen emission spectrum.

Answer 1

Balmer-Rydberg equation

wave number = 1/l = 109737((1/n1^2)-(1/n2^2)) cm-1 (n2>n1)

Paschen series :

the series of lines in the spectrum of the hydrogen atom which corresponds to transitions between the state with principal quantum number n = 3 and successive higher states.

i) longest wavelength(l) , n1 = 3 , n2 = 4

    (1/l) = 109737((1/3^2)-(1/4^2))

     l = 1.87*10^-4 cm

       = 1870 nm

ii) shortest wavelength lines, n1 = 3 , n2 = infinity

(1/l) = 109737((1/3^2)-(1/infinity^2))

    l = 8.2*10%-5 cm

      = 820 nm

      = IR-region

c) wave number = 1/l = 109737((1/n1^2)-(1/n2^2)) cm-1

Balmer series , n1 =2 , n2 = ?

   l = 434 nm

(1/(434*10^-7)) = 109737((1/2^2)-(1/n2^2))

n2 = 5