The maximum kinetic energy of photoelectron is doubled when the wavelength of light incident on the photosensitive changes from `lambda_1` to `lambda_2`. Deduce expression for the threshold wavelength and work function for metals in terms of `lambda_1` and `lambda_2`

The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from lamda_1 " to " lamda_2 The work function of the metal is

Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is lambda . If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is

Let K_(1) be the maximum kinetic energy of photoelectrons emitted by a light of wavelength lambda_(1) and K_(2) corresponding to lambda_( 2). If lambda_(1) = 2lambda_(2) , then

The threshold wavelength for two photosensitive surfaces A and B are lambda 1 and lambda 2 respectively. What is the ratio of the work functions of the two surfaces?

If K_(1) and K_(2) are maximum kinetic energies of photoelectrons emitted when light of wavelength lambda_(1) and lambda_(2) respectively are incident on a metallic surface. If lambda_(1)=3lambda_(2) then

If K_(1) and K_(2) are maximum kinetic energies of photoelectrons emitted when light of wavelength lambda_(1) and lambda_(2) respectively are incident on a metallic surface. If lambda_(1)=3lambda_(2) then

Maximum kinetic energy of photoelectrons emitted from a metal surface, when light of wavelength lambda is incident on it, is 1 eV. When a light of wavelength lambda/3 is incident on the surfaces, maximum kinetic energy becomes 4 times. The work function of the metal is

A dye absorbs a photon of wavelength lambda and re - emits the same energy into two phorons of wavelengths lambda_(1) and lambda_(2) respectively. The wavelength lambda is related with lambda_(1) and lambda_(2) as :

In Young's double-slit experiment, two wavelengths of light are used simultaneously where lambda_(2) = 2 lambda_(1) . In the fringe pattern observed on the screen,

Wavelength of light used in an optical instrument are lambda_(1) = 4000 Å and lambda_(2) = 5000Å then ratio of their respective resolving powers (corresponding to lambda_(1) and lambda_(2) ) is