The electric field of light wave is given as `vec(E )=10^(-3)cos((2pi x)/(5xx10^(-7))-2pixx6xx10^(14)t)hat(x)(N)/(C )` This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is : Given E (in eV) `= (12375)/(lambda("in"Å))`
The electric field of light wave is given as vec(E)=10^(-3)cos((2 pi x)/(5xx10^(-7))-2pi xx 6xx10^(14)t) hat(x) (N)/(C) . This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is: Given, E (in eV) = (12375)/(lambda ( "in" Å ))
Light of energy 2.0 eV falls on a metal of work function 1.4 eV . The stopping potential is
Light of frequency 10^(15) Hz falls on a metal surface of work function 2.5 eV. The stopping potential of photoelectrons in volts is
Light of waelength 4000Å is incident on a metal plate with work function 2eV. The max. K.E. of the photoelectrons is
Electric field associated with a light wave is given E= E_(0) sin [1.57 xx10^(15)t +6.28 xx10^(15) t] V/m. If this light incident on a surface of work function 2.0 eV the stopping potential will be -
The magnetic field associated with a light wave is given, at the origin, by B=B_(0)[sin(3.14xx10^(7))ct +sin(6.28xx10^(7))ct]. If this light falls on a silver plate having a work function fo 4.7 eV, what will be the maximum kinetic energy of the photo electrons? (c=3xx10^(8)ms^(-1),h=6.6xx10^(-34)J-s)
The wave function (in SI unit) for a light wave is given as Psi(x,t) = 10^(3) pi(3 xx 10^(6) x - 9 xx 10^(14)t) . The frequency of the wave is equal to
