A hydrogen atom rises form its n=1 state to the n=4 state by absorbing energy. If the potential energy of the atom in the n=1 state be -13.6 eV, then calculate
(a) potential energy in the n=4 state,
(b) energy obserbed by the atom in transition,
(c) Wavelength of the emitted radiation if the atom returns to its original state.

Answer

`-0.85eV, 12.75 eV ; 970Å`

Explanation

Here, `E_(1)=-13.6eV , n=4, E_(4)=?`
(a) `E_(4)=-(E_(1))/(4^(2))=(-13.6 eV)/16=-0.85eV`
(b) Energy absorbed by the atom in transition form n=1 to n=4 is
`DeltaE=E_(4)-E_(1)=-0.85-(-13.6)`
`=12.75eV`
(c) `DeltaE=(hc)/lambda`
or `lambda=(hc)/(DeltaE)=(6.6xx10^(-34)xx3xx10^(8))/(12.75xx1.6xx10^(-19))`
`=970xx10^(-10)m=970Å`