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Q 12.4  A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

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Frequency of radiation consisting of photons of energy E is given by

 \nu =\frac{E}{h}

E=2.3 eV

Plank's constant(h)=6.62\times10-34 Js

\nu =\frac{2.3\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}

\nu =5.55\times 10^{14}\ Hz

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