Ncert Exercises & Solutions

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?

Hint: The energy of the radiation equals the difference in energy levels.

Step 1: Relate energy with the frequency.
Separation of two energy levels in an atom,

E = 2.3 e V = 2.3 \times 1.6 \times 10^{- 19} = 3.68 \times 10^{- 19} J

Let \(\nu\) 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 =\(h\nu\)

where, h = plank's constant =

6.62 \times 10^{- 34} Js

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Step 2: Find the frequency of the radiation.
\(\nu=\frac{E}{h}=\frac{3.68 \times 10^{-19}}{6.62 \times 10^{-32}}=5.55 \times 10^{14} \mathrm{~Hz}\)
Hence, the frequency of the radiation is

5.6 \times 10^{14} H z .

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