Find the maximum kinetic enegry of photoelectrons liberated form the surface of lithium by electromagnetic radiation whose electric component varies with time as `E = a(1+cos omegat) cos omega_(0)t`, where `a` is a constant, `omega = 6.10^(14)s^(-1)` and `omega_(0) = 360.10^(15)s^(-1)`.(Work function of Lithiuim = 2.39eV)
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`E=a (1+cos omegat)cos omega_(0)t=acosomega_(0)t+a cosomega t cosomega _(0)t` `implies E=a cosomega_(0)t+(1)/(2)acos (omega+omega_(0))t +(1)/(2)acos (omega-omega_(0))t` This is a complex vibration consisting of harmonic components of frequencies `omega_(0),(omega+omega_(0))` and `(omega-omega_(0))`.The highest angular frequency is `(omega+omega_(0))`. now, `hv =phi +K _(max)` so, `K_(max)=(h)/2pi)((omega+omega_(0))-phi`=`(6.6xx 10^(-34))/(2pi)(6xx10^(14)+3.6xx10^(15))-2.39xx1.6xx10^(-19)`=`4.41xx10^(-19)-3.82xx10^(-19)`=`0.59xx10^(-19)`J=0.37eV
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