A difference of 2.3 eV separates two energy levels in an atom. Whatis the frequency of radiation emitted when the atom make atransition from the upper level to the lower level?

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Solution

Given: The energy for separation of two energy levels is 2.3 eV .

The energy difference between two levels is given as,

ΔE=hν

Where, h is Plank’s constant and ν is the frequency of radiation emitted.

By substituting the values in the above equation, we get

ν= ( 2.3 eV )( 1.6× 10 −19 J 1 eV ) 6.62× 10 −32 Js ≈5.55× 10 14 Hz

Thus, the frequency of radiation emitted is 5.55× 10 14 Hz.

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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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