Question:
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?
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?
Updated On: Sep 30, 2023
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Solution and Explanation
Seperation of two energy levels in an atom,
E=2.3eV
\(=2.3×1.6×10^{-19}\)
\(=3.68×10^{-19}J\)
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\(=6.62×10^{-34}Js\)
\(∴v=\frac{E}{h}\)
\(=\frac{3.68×10^{-19}}{6.62×10^{-32}}=5.55×10^{14}Hz\)
Hence,the frequency of the radiation is \(5.6×10^{14}Hz\).
E=2.3eV
\(=2.3×1.6×10^{-19}\)
\(=3.68×10^{-19}J\)
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\(=6.62×10^{-34}Js\)
\(∴v=\frac{E}{h}\)
\(=\frac{3.68×10^{-19}}{6.62×10^{-32}}=5.55×10^{14}Hz\)
Hence,the frequency of the radiation is \(5.6×10^{14}Hz\).
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