A metal surface does not emit any electron if the energy of incident radiation is less than the photoelectric work function of the metal - justify the statement mathematically.

Why are the photoelectric work functions different for differet metals?

Write down the relation between threshold frequency and photoelectric work function for a metal.

Statement I: The stopping potential becomes double when the frequency of the incident radiation is doubled Statement II: Work function of the metal and the threshold frequency of photoelectric effect are proportional to each other.

When radiation of wavelength 4940 Å is incident on a metal surface photoelectricity is generated. For a potential difference 0.6 V between cathode and anode, photocurrent stops. For another incident radiation, the stopping potential changes to 1.1V. Find work function of the metal and wavelength of the second radiation. (h = 6.6 xx 10^(-27) " erg".s, e = 1.6 xx 10^(-19) C)

When light incident on a metal has energy less than the work function of the metal, then no electron is emiited from the surface of the metal. Mathematically justify this statement.

If the intensity of incident radiation on a metal surface is doubled, what happens to the kinetic energy of the electrons emitted?

Work function of a metal is 4.0 eV. Find the maximum value of wavelength of a radiation that can emit photoelectrons from the metal.

Energy of incident photon on a metal surface is 5.0 eV and photoelectric stopping potential is 3.23V. Determine the threshold wavelength for the metal.

Statement I: The maximum velocity of the photoelectron emitted from a metal surface does not increase even if the intensity of the incident electromagnetic wave is increased. Statement II: Einstein's photoelectric equation: (1)/(2) mv_("max")^(2) = hf - W_(0)

thresold Wavelength of incident light on a metal surface is 404 Å . Determine the work function.