In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits is given by ri:rf=16:4. The wavelength of photon emitted due to this transition is nm. (Given Rydberg constant = 1.0973 107 /m)
- 121
- 242
- 486
- 974
Answer
(3) 486
Solution
Bohr radius: rn=0.529 n2 Å ⇒ r n2
⇒ ri:rf=ni2:nf2=16:4=42:22 ⇒ ni=4, nf=2 (Balmer series)
1 =R ( 1nf2- 1ni2 )=R ( 14- 116 )= 3R16
⇒ = 163 1.0973 107 m=4.86 10-7 m
∴ 486 nm
Answer: (3)