if the speed of photoelectrons is `10^4` m/s, what should be the frequency of the incident radiation on a potassium metal ? Given : Work function of potassium =2.3 eV

If the speed of photoelectrons is 10^(6) ms^(-1) , what should be the frequency of the incident radiation on the potassium metal ? (Work function of potassium = 2.3 eV).

If the speed of photoelectrons is 10^(4) ms^(-1) , what should be the frequency of incident radiation on the potassium metal ? Work function of potassium = 2.3 eV.

What is the energy of emitted photoelectrons is light of frequency 10^(16) Hz is incident on a sodium target ? Work function of sodium is = 2.5 eV.

The work function of a metal is 3.4 eV. If the frequency of incident radiation is increased to twice, then the work function of the metal becomes.

The work function of a metal is 3.4 eV. If the frequency of incident radiation is increased to twice, then the work function of the metal becomes.

Radiation of intensity 0.5 xx 10^-4 W//m^2 falls on the emitter in a photoelectric set-up. The emitter (cathode) is made up of potassium and has an area of 5 cm^2 . Let us assume that the electrons from only the surface are knocked off by the radiation. According to the wave theory,what will be the time required to notice some deflection in the microammeter connected in the circuit? (Given the metallic radius of potassium atom is 230 pm and work function of potassium is 2.3 eV.)