AP EAMCET202125 Aug 2021Evening ShiftChemistryStructure of AtomActual
If the wavelength of the first line in Balmer series is 656 nm, then the wavelength of its second line and limiting line respectively are
Options
- A485.9 nm and 434 nm
- B485.9 nm and 364.4 nm
- C715 nm and 434 nm
- D608 nm and 415.2 nm
Correct answer
B. 485.9 nm and 364.4 nm
Step-by-step solution
According to H-spectrum; Rydberg formula 1 =R_ H ( 1 n₁^2 - 1 n₂^2 ) For first line, n₁=2, n₂=3 (Balmer series) =656 ~nm aligned 1 656 & =R_ H [ 1 2^2 - 1 3^2 ] & = 5 36 R_ H ...(i) aligned For second line, n₁=2 and n₂=4 aligned 1 & =R_ H [ 1 2^2 - 1 4^2 ] & = 3 16 R_ H ...(ii) aligned Dividing Eqs. (i) and (ii) we get, aligned 656 & = 5 36 16 3 & =485.9 ~nm aligned Similarly, wavelength of limiting line i.e. n₂= and n₁=2 aligned 1 & =R_ H [ 1 2^2 - 1 ^2 ]= R_ H 4 = 109737 4 & =364.4 ~nm aligned Hence, wavelength o