Light of frequency $7.21 \times 10^{14} \; Hz$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \; m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

The work-function of a metal is $1 eV$. Light of wavelength $3000 \text{Å}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

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:

In photoelectric effect, the graph of stopping potential $(V_0)$ versus frequency $(\nu)$ is a straight line. The slope of this graph is . . . . . . .

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

$A$ 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.