One milliwatt of light of wavelength 4560 A is incident on a cesium surface. Calculate the photoelectric current liberated assuming a quantum efficiency of 0.5 %. Given Planck's constant `h=6.62xx10^(-34)J-s` and velocity of light `c=3xx10^(8)ms^(-1)`.

A photon has a wavelength of 3.5Å. Calculate its mass (Plank's constant, = 6.626 xx 10^(-34) Js . Velocity of light = 3 xx 10^(8) ms^(-1) )

The energy of a photon wavelength k = 1 meter is (Planck's constant = 6.625 xx 10^(-34) Js, speed of light = 3 xx 10^(8) m//s )

For photoelectric emission, threshold wavelength of a metal is 2460 Å . If ultraviolet light of wavelength 1810 Å be incident on the metal, what will be the maximum kinetic energy of the emitted electrons? [Planck's constant = 6.62xx10^(-34) J.s , velocity of light in veccuum = 3xx10^(8) m.s^(-1) ]

Calculate the energy in joule corresponding to light of wavelength 45nm (Planck's constant h = 6.63 xx 10^(-34)Js , speed of light c = 3 xx 10^(8)ms^(-1) )

Calculate the wavelength in nanometers of a quantum of light with frequency of 8times10^(15)s^(-1) .The given value of Planck's constant is 6.63times10^(-34)Js .The velocity of light is 3.0times10^(8)ms^(-1)