Question

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Answer 1

Separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10⁻¹⁹

= 3.68 × 10⁻¹⁹ J

Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level.

We have the relation for energy as:

E = hv

Where,

h = Planck’s constant = 6.62 × 10⁻³⁴ Js

∴ v = `"E"/"h"`

= `(3.68 xx 10^(-19))/(6.62 xx 10^(-32))`

= 5.55 × 10¹⁴ Hz

Hence, the frequency of the radiation is 5.6 × 10¹⁴ Hz.