1.

Wavelength of radiation emitted when electron in hydrogen atom transist from 4th energy level to 2nd energy level

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



Discussion

No Comment Found

Related InterviewSolutions