The electric field of an electromagnetic wave changes with time as E=K(1+cosOmega t) cos omega t, where Omega= 5xx10^(15) s^(-1), omega= 2xx10^(16) s^(-1) and K is constant. This radiation is incident on a sample of hydrogen atoms initially in ground state. Assume that atoms absorb light as photons. neglecting recoil of hydrogen nucleus on ionisation, what will be the energy of ejected electrons from hydrogen. [The ionisation energy of hydrogen atom= 13.6 eV and h=2pi xx 6.6xx 10^(-16) eV-s]

The electric field of an electromagnetic wave changes with time as E=K

1.	The electric field of an electromagnetic wave changes with time as E=K(1+cosOmega t) cos omega t, where Omega= 5xx10^(15) s^(-1), omega= 2xx10^(16) s^(-1) and K is constant. This radiation is incident on a sample of hydrogen atoms initially in ground state. Assume that atoms absorb light as photons. neglecting recoil of hydrogen nucleus on ionisation, what will be the energy of ejected electrons from hydrogen. [The ionisation energy of hydrogen atom= 13.6 eV and h=2pi xx 6.6xx 10^(-16) eV-s]
Answer» 0.7eV 0.9 eV 1.4 eV 2.9 eV Solution :The expression for electric field can ALSO be written as `E=K cos omega t+1/2 K cos(omega-Omega) t+1/2 K cos (omega+Omega) t` The three terms CORRESPOND to photons of energies `h omega, h(omega-Omega)` and `h(omega-Omega) ` . The LATTER exceeds the ionisation by 2.9 eV . that DIFFERENCE equals ejected electrons energy.