A source S₁ is producing, 10¹⁵ photons per second of wavelength 5000 Å. Another source S² is producing 1.02 × 10¹⁵ photons per second of wavelength 5100 Å Then,(power of S²) to the (power of S¹) is equal to :

For intensity I of a light of wavelenght 5000 Å the photoelectron saturation current is 0.40 µA and stopping potential is 1.36 V, the work function of metal is

The work function of aluminium is 4.2 eV. If two photons,each of energy 3.5 eV strike an electron of aluminium, then emission of electrons

The frequency and work function of an incident photon are v and φ₀. If v₀ is the threshold frequency then necessary condition for the emission of photoelectron is

X-rays are produced in X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell, the work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero the voltage of the anode relative to the cathode must be made

A photon of 1.7 × 10^–13 joule is absorbed by a material under special circumstances. The correct statement is

The stopping potential (V₀) versus frequency (v) plot of a substance is shown in figure, the threshold wavelength is

All electrons ejected from a surface by incident light of wavelength 200nm can be stopped before travelling 1m in the direction of uniform electric field of 4N/C. The work function of the surface is

An X-ray tube is operated at 15 kV. Calculate the upper limit of the speed of the electrons striking the target.