Ultraviolet light of wavelength `800 A` and `700 A` when allowed to fall on hydrogen atoms in their ground states is found to liberate electrons with kinetic energies `1.8eV` and `4.0eV`, respectively. Find the value of Planck's constant.

Answer

`6.57xx10^(-34)Js`

Explanation

K.E. of photoelectrons, `K=(hc//lambda)-phi_(0)`,
case (i), `K=1.8xx1.6xx10^(-19)J`,
`lambda=800xx10^(-10)m`,
case (ii), `K=4.0xx1.6xx10^(-19)J`,
`lambda=700xx10^(-10)m`
`:. 1.8xx1.6xx10^(-19)=(hxx3xx10^(8))/(800xx10^(-10))-phi_(0)........(i)`
`4.0xx1.6xx10^(-19)=(hxx3xx10^(8))/(700xx10^(-10))-phi_(0)........(ii)`
Subtracting (i) from (ii) we get
`(4.0-1.8)1.6xx10^(-19)=hxx3xx10^(16)[1/7-1/8]`
on solving we get, `h=6.57xx10^(-34) Js`