`U.V`. light of wavelength `800A^(@)&700A^(@)` falls on hydrogen atoms in their ground state `&` liberates electrons with kinetic energy `1.8eV` and `4eV` respectively. Calculate planck's constant.

Answer

`6.57xx10^(34)Js`

Explanation

`(hc)/(lambda_(1))=w_(0)+E_(1)" "(hc)/(lambda_(2))=w_(0)+E_(2)`
`(hc)/(lambda_(1))-E_(1)=w_(0)" ".....(i)`
`(hc)/(lambda_(2))-E_(2)=w_(0)" ".....(ii)`
`(hc)/(lambda_(1))-E_(1)=(hc)/(lambda_(2))-E_(2)`