Question

The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quantum number. Imagine a situation where the electron makes a transition from n = 3 to n = 1 state. Where does the photon emitted in this transition lie in the electromagnetic spectrum? (A) visible region (B) Infra-red region (C) Ultra-violet region (D) X-ray region

Answer 1

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

$E_3 =- \frac{13.6 }{3^2}$

$E_3 =- \frac{13.6 }{9}$ = -1.511 eV

Energy of the orbit where n = 1 is given as follows

$E_1 =- \frac{13.6 }{1^2}$

$E_1 =- \frac{13.6 }{1}$ = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .