Question

1. Two energy levels of an electron in hydrogen atom are separated by 2.55 eV. Find the wavelength of radiation emitted when the electron makes transition from the higher energy level to the lower energy level.
2. In which series of hydrogen spectrum this line shall fall?

Answer 1

a. Given, E = 2.55 eV = 2.55 × 1.6 × 10⁻¹⁹ J

E = 4.08 × 10⁻¹⁹ J

Since, `E = (hc)/lambda`

⇒ `4.08 xx 10^-19 = (6.626 xx 10^-34 xx 3 xx 10^8)/lambda`

λ = 4872 Å

b. This line falls in the Balmer series of the hydrogen spectrum.