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NCERT Class XII Chapter
Atoms
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Question : 4 of 18
Marks: +1, -0
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?
Solution:  
Here E = 2.3 eV = 2.3 × 1.6 × 10−1910^{-19} J
As E = hÏ…
∴ Frequency , υ = Eh\frac{E}{h} = 2.3×1.6×10−196.6×10−34\frac{2.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 5.6 × 101410^{14} Hz
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