Atomic StructureEasy
Question
Wavelength of photon which have energy equal to average of energy of photons with λ1 = 4000 Å and λ2 = 6000 Å will be
Options
A.5000 Å
B.4800 Å
C.9600 Å
D.2400 Å
Solution
$E = \frac{1}{2}\left( E_{1} + E_{2} \right) \Rightarrow \frac{1}{\lambda} = \frac{1}{2}\left( \frac{1}{4000} + \frac{1}{6000} \right)$
$\therefore\lambda = 4800\overset{o}{A}$
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