The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in hydrogen atom will be (Given ionization energy of H = 2.18 10-18 J atom-1 and h = 6.625 10-34 J s)
- 1.54 1015 s-1
- 1.03 1015 s-1
- 3.08 1015 s-1
- 2.00 1015 s-1
Answer
(3) 3.08 1015 s-1
Solution
As per NCERT, En = - 2.18 10-18n2 J
E = 2.18 10-18 ( 112 - 142 )
= 2.18 10-18 1516
= 2.044 10-18 J
∵ E = h
∴ = 2.044 10-186.625 10-34
= 3.08 1015 s-1
Answer: (3)