(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

Answer

(i) `913Å` (ii) `10.2 eV`

Explanation

(i) `E_(n)=(13.6)/(n^(2))` ( for `H`-atom)
`E=13.6 eV` (Ground state)
Ionisation energy `=13.6 eV`
Max. wave length req.
`13.6xx1.6xx10^(-19)=(hc)/(lambda_(max))=(6.6xx10^(-34)xx3xx10^(8))/(lambda_(max))rArr lambda_(max)913Å`
(ii) First excited state `n=2`
total energy of electron
`:. E_(n=2)=(13.6)/(4)eV= -3.4eV`
`:.` So work function `phi=3.4 eV`
maximum Kinetic energy`=`Energy incident-work function
`rArr 13.6-3.4=10.2 eV`