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Q.

A difference of 2.3 eV separates two energy levels in an atom. The frequency of radiation emitted when an atom makes transition from the upper energy level to the lower level corresponding to above mentioned energy levels is

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a

5.6 x 1014 Hz

b

6.5 x 1012 Hz

c

6.5 x 1013 Hz

d

5.6 x 1013 Hz

answer is B.

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Detailed Solution

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ΔE=hv⇒v=2.3×1.6×10−196.625×10−34Hz=5.57×1014Hz
≈5.6 x 1014 Hz

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