Photoemission from metal surface takes place for frequencies '$$v_1$$' and '$$v_2$$' of incident rays $$\left(v_1>v_2\right)$$. Maximum kinetic energy of photoelectrons emitted is in the ratio $$1: \mathrm{K}$$. The threshold frequency of metallic surface is
A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)
Light of frequency two times the threshold frequency is incident on photosensitive material. If the incident frequency is made $$\left(\frac{1}{3}\right)^{\text {rd }}$$ and intensity is doubled, then the photoelectric current will
On a photosensitive surface, if the intensity of incident radiation is increased, the stopping potential