Question

Electronically excited hydrogen emits in the visible part of the spectrum in a series of lines known as the Balmer series. Each of these transitions terminates in the n=2 level of hydrogen. What is the energy and wavelength and upper state quantum number for the first four of these transitions starting with the longest wavelength emission?

Answer 1

n1 n2 (upper state quantum number) wavelength (nm) Energy (J)

2 3 656 3.03 x 10^-19

2 4 486 4.09 x 10^-19

2 5 434 4.58 x 10^-19

2 6 411 4.84 x 0^-19

1) Wavelength can be calculated by using the formula

1/λ = R[ 1/n1^2- 1/n2^2]

λ = wavelength

R = Rydberg constant = 1.097 x 10⁷ m^-1

For 2 to3 transition, n1=2 and n2=3

   1/λ = R[ 1/n1^2- 1/n2^2]

1/λ = (1.09x 10^7 m-1) ( 1/2^2 -1/3^2)

  1/λ = (1.09x 10^7 m-1)  x (5/36)

  λ = (36/5) ( 1/1.09x 10^7 m-1 )

   = 6.56 x 10^-7 m

   = 655 nm

Therefore , λ = 656 nm

Energy for 2 to 3 transition:

E = hc/ λ

= (6.626 x 10^-34 J.s) (3 x 10⁸ m/s) / (656 x 10^-9 m)

= 3.03 x 10^-19 J

E =  3.03 x 10^-19 J

Similarly, wavelengh and energy can be calculated for other transitions also.