(a) Find the maximum wavelength `lambda_90)` of light which can ionize a hydrogen atom in its ground (b) light of wavelength `lambda_(0)` is inclined on ahydrogen atom which is in its first excited state find the kinrtic energy of the electron coming out

Explanation

(a)To ionized a hydrogen atom inn ground state a minimum of `13.6eV` energy should be to itA photon of light should have this much of energy in order to lonine a hydrogen atom Thus
`(hc)/(lambda_(0)) = 13.6eV`
`or, `lambda_(0) = (1242eVnm)/(13.6eV)= 91.3nm`
(b) The energy of the hydrogen atom in the first excited state is `(13.6eV)/(4) = 9.4eV`Thus `3,4eV` of energy is needed to take the electron out of the energy of a photon of the light of wavelength `lambda_(0)is 13.5eV` Thus the electron coming out will have a kinetic energy `13.6eV- 3.4eV`