Positive rays were discovered by

An ionisation chamber, with parallel conducting plates as anode and cathodes has $5 imes {10^7}{\rm{\;c}}{{\rm{m}}^{ - 3}}$ electrons and the same number of singly charged positive ions per ${\rm{c}}{{\rm{m}}^3}$. The electrons are moving toward the anode with velocity 0.4${\rm{m}}{{\rm{s}}^{ - 1}}$. The current density from anode to cathode is $4{\rm{\mu A}}{{\rm{m}}^{ - 2}}.$ The velocity of positive ions moving towards cathode is

Positive rays consists of

If the work function of a photometal is $6.825\;eV.$ Its threshold wavelength will be $\left( {c = 3 imes {{10}^8}\;m/s} \right)$

Planck’s constant has the dimensions of

When yellow light is incident on a surface, no electrons are emitted while green light can emit. If red light is incident on the surface, then

Lenard observed that when light below a certain frequency is incident on a metal, no photoelectrons are emitted. This frequency is called the:

A photosensitive material is illuminated with light of frequency $2.5

u_0$, where$

u_0$ is the threshold frequency. What will be the photoelectric current if the intensity of the incident light is reduced to half while the frequency remains the same?

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 imes {10^{ - 3}}$ watt will be $(h = 6.6 imes {10^{ - 34}}{\rm{Js)}}$

The variation of photoelectric current given by the photocell, with the intensity of light, is give by a graph, which is a straight line with

Light of frequency ${\rm{1}}{\rm{.5}}$ times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled ?