Chemistry, asked by subi20qhpcmmfi, 9 months ago

In hydrogen atom, an electron jumps from 3rd orbit to the 2nd orbit. Calculate the wavelength, frequency and energy of the radiation emitted

Answers

Answered by harshkrishnan
3

Answer:

6564Å

Solution :

En=−21.8×10−19n2Jatom−1

ΔE=E3−E2=21.8×10−19(122−132)=3.03×10−19J

ΔE=hv=hcλ

λ=hcΔE=(6.63×10−34Js)(3×108ms−1)(3.03×10−19J)=6.564×10−7m=6564×10−10m=Å

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