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
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
Similar questions