The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

Given that a photon of light of wavelength $10,000\;\mathring A$ has an energy equal to $1.23\; eV$. When light of wavelength $5000\;\mathring A$ and intensity $I_0$ falls on a photoelectric cell,the saturation current is $0.40 \times 10^{-6}\; A$ and the stopping potential is $1.36\; V$; then the work function is ..... $eV$.

$A$ mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission,as it provides a number of spectral lines ranging from the $UV$ to the red end of the visible spectrum. In our experiment with a rubidium photocell,the following lines from a mercury source were used:
$\lambda_1 = 3650 \,\mathring{A}, \lambda_2 = 4047 \,\mathring{A}, \lambda_3 = 4358 \,\mathring{A}, \lambda_4 = 5461 \,\mathring{A}, \lambda_5 = 6907 \,\mathring{A}$
The stopping voltages,respectively,were measured to be:
$V_{01} = 1.28 \,V, V_{02} = 0.95 \,V, V_{03} = 0.74 \,V, V_{04} = 0.16 \,V, V_{05} = 0 \,V$
Determine the value of Planck's constant $h$,the threshold frequency,and the work function for the material.

$A$ photosensitive metallic surface is illuminated alternately with lights of wavelength $3100 \mathring A$ and $6200 \mathring A$. It is observed that the maximum speeds of the photoelectrons in the two cases are in the ratio $2:1$. The work function of the metal is $(hc = 12400 \, eV \mathring A)$.

Light of frequency $\nu$ is incident on a substance of threshold frequency $\nu_0$ (where $\nu_0 < \nu$). The maximum kinetic energy of the emitted photo-electron will be:

Given below are two statements:
Statement-$I$: The figure shows the variation of stopping potential $(V_0)$ with frequency $(v)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\frac{h}{e}$,where $h$ is Planck's constant and $e$ is the charge of an electron.
Statement-$II$: $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation having the same frequency.
In the light of the above statements,choose the most appropriate answer from the options given below.