$A$ photon of energy $5.5 \ eV$ strikes a surface that emits photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping potential for these electrons is ............ $V$.

The threshold frequency of cesium is $5.16 \times 10^{14} \ Hz$. Then its work function is . . . . . . $eV$.

The threshold frequency of potassium is $3 \times 10^{14} \ Hz$. The work function is ...... .

Light of frequency $4v_0$ is incident on a metal surface with a threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is:

Cut-off potentials for a metal in the photoelectric effect for light of wavelengths $\lambda_1$,$\lambda_2$,and $\lambda_3$ are found to be $V_1$,$V_2$,and $V_3$ volts. If $V_1$,$V_2$,and $V_3$ are in Arithmetic Progression,then $\lambda_1$,$\lambda_2$,and $\lambda_3$ will be in:

$A$ 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.