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Wavelength of radiation emitted when electron in hydrogen atom transist from 4th energy level to 2nd energy level |
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Answer» Answer: 486.27 nm Explanation: energy of 4TH shell =13.6/(4)^2=0.85 eV energy of 2nd shell =13.6/(2)^2=3.4 eV so energy emmited =3.4-0.85=2.55 eV E=hc/lambda so lambda=hc/E=1240/2.55=486.27 nm please mark my answer as brainliest and FOLLOW me |
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