What will be the frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a H atom. A) 4.57 * 10 ^ 14 ^ 4 B)-l - 4.57 * 10 ^ 14 c) 3.57 * 10 ^ 14 d) 4.57 * 10 ^ 4
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Step-by-step explanation:
In hydrogen spectrum the spectral lines are expressed are expressed in term of wave number v⃗ obey the following formula ltbr. Wave number v⃗ =RH(1n2i−1n2f) (where RH= Rydberg constant 109677cm−1)
v⃗ =109677cm−1(122−132)
v⃗ =109677×536=15232gcm−1
v⃗ =1λ
or λ=1v=115232g=6.564×10−5cm
Wavelength, λ=6.564×10−7m ltbr. Energy E=hcλ
=6.626×10−34Js×30×108ms−16.564×10−7m
=3.028×10−19J
Frequency v=cλ=cλ=3.0×108ms−16.564×10−7m
=0.457×1015s−1=4.57×1014s−1
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