When ultraviolet light of wavelengths 800 Å and 700 Å are incident on the hydrogen atom at ground state electrons are emitted with energies 1.8 eV and 4 eV, respectively. Determine the value of Planck's constant.

Explanation

Let the ground state energy of hydrogen atom `=-E_0`.
Hence , the minimum amount of energy `E_0` is required to liberate its electron, i.e., the work function of hydrogen atom,
`W_0 =E_0`. So, if the incident photon can provide E amount of kinetic energy to the electron , then
`hf=E+E_0 or, (hc)/(lambda)=E+E_0" " .....(1)`
In the first case,
`lambda_1=800 "Å" =800xx10^(-8) cm`,
`E_1=1.8 eV=1.8xx1.6xx10^(-12) erg`
In the second case
`lambda_2=700 "Å"=700xx10^(-8)` ,
`E_2=4.0 eV =4.0xx1.6xx10^(-12) erg`
From equation (1) , we get
`(hc)/(lambda_1)-(hc)/(lambda_2)=E_1-E_2 or, he (1/lambda_1-1/lambda_2)=E_1-E_1 or, h=(E_2-E_1)/(c)*(lambda_1lambda_2)/(lambda_1-lambda_2)`
`therefore h=((4.0-1.8)xx1.6xx10^(-12))/(3xx10^(10))xx(800xx700xx10^(-16))/((800xx700)xx10^(-8))=6.57xx10^(-27)erg*s`