The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

When a point source of light is at a distance of $1 \ m$ from a photocell,the cut-off voltage is found to be $V$. If the same source is placed at $2 \ m$ distance from the photocell,the cut-off voltage will be

$A$ beam of light falls on a metal surface such that photo-electrons are generated. If the power of the light source starts to decrease linearly with time $t$, then the variation of the photocurrent $I$ and the magnitude of the stopping potential $|V|$ with time is best represented by:

$A$ 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.

The work function of a metal is $3 \ eV$. The color of the visible light that is required to cause emission of photoelectrons is

$A$ beam of electromagnetic radiation of intensity $6.4 \times 10^{-5} \; W/cm^{2}$ is comprised of wavelength $\lambda = 310 \; nm$. It falls normally on a metal surface (work function $\varphi = 2 \; eV$) of surface area $1 \; cm^{2}$. If one in $10^{3}$ photons ejects an electron, the total number of electrons ejected in $1 \; s$ is $10^{x}$. Then $x$ is: $(hc = 1240 \; eV \cdot nm, 1 \; eV = 1.6 \times 10^{-19} \; J)$