The variation of stopping potential for metals $A$,$B$,$C$ and $D$ with the frequency of incident radiation is shown in the figure. For which metal is the stopping potential higher for a given frequency of incident radiation $(v)$ if the threshold frequency is lower $(\nu_o)$?

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} \, m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} \, m$ is (Given $h = 4.14 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

If the frequency of incident radiation $(v)$ is increased,keeping other factors constant,the stopping potential ($v > v_0$,threshold frequency) will:

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:

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

The variation of stopping potential $(V_0)$ as a function of the frequency $\nu \ (\times 10^{14} \ Hz)$ of the incident light for a metal is shown in the figure. The work function of the surface is $........... \ eV$.