# 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? - Physics

Numerical

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?

#### Solution

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.

Concept: Energy Levels
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 12 Atoms
Exercise | Q 12.4 | Page 436
NCERT Class 12 Physics Textbook
Chapter 12 Atoms
Exercise | Q 4 | Page 436

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