Question

Calculate the de-Brogile wavelength of the elctron orbiting in the `n=2` state of hydrogen atom.

Answer 1

Energy in orbit for which `n=2`.
`E=(-13.6)/(2^(2))eV=-3.4eV`
Now, `3.4xx1.6x10^(-19)=(hc)/(lambda)implies=(6.6xx10^(-34)xx3xx10^(8))/(3.4xx1.6xx10^(-19))=3.639xx10^(-7)=3639Å`.