In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is:

The threshold wavelength for photoelectric emission from a metal surface is $5200 \ \mathring A$. Which of the following sources will cause photoelectric emission from this surface?

With reference to the observations in photoelectric effect,identify the correct statements from below:
$A.$ The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.
$B.$ The value of saturation current increases on moving the source of light away from the metal surface.
$C.$ The maximum kinetic energy of photoelectrons decreases on decreasing the power of $LED$ (light emitting diode) source of light.
$D.$ The immediate emission of photoelectrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.
$E.$ Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves.
Choose the correct answer from the options given below:

Photoelectrons are emitted from a metal surface for frequencies $\nu_1$ and $\nu_2$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1 : k$,then the threshold frequency of the metal surface is: