The maximum K.E of elecrons emitted by a photocell is 3eV. What is the stopping potential?

Q. The maximum energy of photo electrons emitted in a photocell is 2 eV. For no photo-electrons to reach the anode, the stopping potential should be:

Q. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is $v_0$ and the maximum kinetic energy of the photo electrons is $K_{max}$ . When the ultraviolet light is replaced by X-rays, $K_{max}$increases but $v_0$ , decreases.
Statement -2: Photo electrons are emitted with speeds ranging from zero to a maximum value.Below a certain negative voltage $V_0$ ,no photons are emitted in a photocell

Q. A photocell emits $4\times {10}^{12}$ electrons per second with maximum KE of E when the source of light is at a distance d from the photocell. If the distance is made equal to 2d, the number of electrons emitted per second and maximum K.E.will be

Q. In an experiment on the photo-electric effect, singly ionized helium is excited electronically to different energy levels. The light emitted by ionized helium is incident on a photo-electric plate in a photocell. When helium is excited to fourth energy level, then the observed stopping potential of the photocell is found to be five times the stopping potential measured when the photoelectrons are produced by using light emitted by hydrogen atom, excited to the third energy level. Determine the work function of the material of the photo-electric plate.

Q. Light of wavelength $4000A^o$ is incident on a metal surface of work function 2.5 eV. Given $h=6.62\times {10}^{-34}Js$, $c=3\times {10}^{8}m/s$, the maximum KE of photo-electrons emitted and the corresponding stopping potential are respectively