JEE Main2012ChemistryStructure of AtomActual
The limiting line in Balmer series will have a frequency of (Rydberg constant, R_ =3.29 10¹⁵ cycles / s )
Options
- A8.22 10¹⁴ ~s ⁻¹
- B3.29 10¹⁵ ~s ⁻¹
- C3.65 10¹⁴ ~s ⁻¹
- D5.26 10¹³ ~s ⁻¹
Correct answer
A. 8.22 10¹⁴ ~s ⁻¹
Step-by-step solution
v = 1 =R_H Z ( 1 n₁^2 - 1 n₂^2 ) In Balmer series n₁=2 & n₂=3,4,5 ... Last line of the spectrum is called series limit. Limiting line is the line of shortest wavelength and high energy when n₂= aligned v = & 1 = R_H n₁^2 = 3.29 10¹⁵ 2^2 = 3.29 10¹⁵ 4 & =8.22 10¹⁴ ~s ⁻¹ aligned