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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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?

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प्रश्न

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?

संख्यात्मक
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उत्तर

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.

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पाठ 12: Atoms - EXERCISES [पृष्ठ ३०५]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 12 Atoms
EXERCISES | Q 12.3 | पृष्ठ ३०५
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 12 Atoms
Exercise | Q 12.4 | पृष्ठ ४३६

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