Ultraviolet radiations of wavelengths 800 Å and 700 Å when allowed to fall on Hydrogen atoms in their ground state, were found to liberate electrons with maximum kinetic energies `1.96` eV and `4.18` eV respectively. Calculate Planck,s constant and work function or ionisation energy of the hydrogen atom.

Explanation

`lambda_(1) = 8xx10^(-8)m, lambda_(2) = 7 xx 10^(-8) m`
`k_(1)=1.96xx1.6 xx 10^(-19) J = 3.136xx10^(-19)J`
`k_(2) = 4.18xx1.6 xx10^(-19) J = 6.688xx10^(-19)J`
We know that `E =w+(KE)_(max)`
i.e., `hv_(1)=w+k_(1)`
`hv_(2) - w + k_(2)`
`therefore (k_(2)-k_(1))=h(v_(2)-v_(1))`
i.e., `h=((k_(2)-k_(1)))/(c(1/lambda_(2)-1/lambda_(1)))`
or `h=((k_(2)-k_(1))(lambda_(1)lambda_(2)))/(c(lambda_(2)-lambda_(1)))=((6.688-3.136)10^(-19)xx8xx7xx10^(-16))/(3xx10^(8)(8-7)10^(-8))`
Planck.s constant `h = 6.630x10^(-34)`Js.
Work function, `=((6.630xx10^(-34)xx3xx10^(8))/(8xx10^(-8)))-(3.136xx10^(19))`
`= 24.86 xx 10^(-19)-3.136 xx 10^(-19)`
`=21.724xx10^(-19)J div 1.6 xx 10^(-19) = 13.58`eV
i.e., Ionisation energy `= 13.58` eV.