Two identical metal plates show photoelectric effect by a light of wavelength `lambda_A` falling on plate A and `lambda_B` on plate B`(lambda_A=2lambda_B)`. The maximum kinetic energy is

Two identical metal plates show photoelectric effect. Light of wavelength lamda_A falls on plate A and lamda_B fall on plate B lama_A=2lamda_B , The maximum KE of the photoelectrons are K_A and K_B , respectively, Which one of the following is true?

Let K_(1) be the maximum kinetic energy of photoelectrons emitted by a light of wavelength lambda_(1) and K_(2) corresponding to lambda_( 2). If lambda_(1) = 2lambda_(2) , then

Let be the maximum kinetic energy of photoelectrons emitted by light of wavelength lambda_(1) and lambda_(2) correspondingto wavelength lambda_(2) . If lambda_(1) = 2lambda_(2) then:

Cut off potentials for a metal in photoelectric effect for light of wavelength lambda_(1), lambda_(2) and lambda_(3) is found to be V_(1), V_(2) and V_(3) volts if V_(1), V_(2) and V_(3) are in Arithmetic Progression and lambda_(1), lambda_(2) and lambda_(3) will be :

Cut off potential for a metal in photoclectric effect for light of wavelength lambda_(1), lambda_(2) and lambda_(3) is found to be V_(1),V_(2) and V_(3) volts , If V_(1),V_(2) and V_(3 are in Arithmetic progression then lambda_(1),lambda_(2) and lambda_(3) will be in

If K_(1) and K_(2) are maximum kinetic energies of photoelectrons emitted when light of wavelength lambda_(2) and lambda_(1) respectively are incident on a metallic surface. If lambda_(1)=3lambda_(2) then

Maximum kinetic energy of photoelectrons from a metal surface is K_0 when wavelngth of incident light is lambda . If wavelength is decreased to lambda|2 , the maximum kinetic energy of photoelectrons becomes (a) =2K_0 (b) gt2K_0 (c ) lt2K_0