In photoelectric effect, the K.E. of electrons emitted from the metal surface depends upon
Intensity of light
Frequency of incident light
Velocity of incident light
Both intensity and velocity of light
Energy of photon whose frequency is ${10^{12}}MHz,$ will be
When wavelength of incident photon is decreased then
A photo sensitive material is at $9\,m$ to the left of the origin and the source of light is at $7\,m $ to the right of the origin along $x$ -axis. The photosensitive material and the source of light start from rest and move, respectively, with $8 \widehat i \, m / s$ and $4 \widehat i \, m / s$ . The ratio of intensities at $t = 0$ to $t = 3\,s$ as received by the photosensitive material is :-
Monochromatic light with a frequency well above the cutoff frequency is incident on the emitter in a photoelectric effect apparatus. The frequency of the light is then doubled while the intensity is kept constant. How does this affect the photoelectric current?
A convex lens of focal length $40 \mathrm{~cm}$ forms an image of an extended source of light on a photoelectric cell. A current I is produced. The lens is replaced by another convex lens having the same diameter but focal length $20 \mathrm{~cm}$. The photoelectric current now is: