(a) Find the maximum wavelength `lambda_(0)` of light which can ionize a hydrogen atom in its ground (b) light of wavelength `lambda_(0)` is inclined on a hydrogen atom which is in its first excited state find the kinetic energy of the electron coming out

Explanation

(a) Ionisation energy of hydrogen atom =13.6 eV
`lambda_(0)=(1242)/(E(eV)) nm=(1242)/(13.6)=91.3 nm`
Energy of hydrogen atom in first excited state (n=2)
`E_(2)=-(13.6)/(n^(2))=-(13.6)/(2^(2))=-3.4 eV`
When a photon of wavelength `lambda_(0)` is incident on it ,
Total energy `=-3.4+13.6=10.2 eV`
Electron goes to `n=oo`
Total energy is in the form of kinetic energy