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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Solution

Difference between energy levels, E_{2} - E_{1} = 2.3 eV = 2.3 \times 1.6 \times 10^{- 19} J F r e q u e n c y o f t h e e m i t t e d r a d i a t i o n i s, v = \frac{E_{2} - E_{1}}{h} \Rightarrow v = \frac{2.3 \times 1.6 \times 10^{- 19}}{6.6 \times 10^{- 34}}

\Rightarrow

v = \frac{3.68 \times 10^{15}}{6.6}

= 0.557 \times 10^{15} Hz = 5.6 \times 10^{14} Hz .

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