Light of wavelength $\lambda$ is incident on the surface of a metal having work function $\phi$, causing the emission of electrons. What is the maximum velocity of the emitted electrons? (Given: $c = \text{velocity of light}$, $h = \text{Planck's constant}$, $m = \text{mass of electron}$)

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is [$h=$ Planck's constant,$m=$ mass of electron and $c=$ speed of light]

$A$ metal plate of area $1 \times 10^{-4} \, m^2$ is illuminated by radiation of intensity $16 \, mW/m^2$. The work function of the metal is $5 \, eV$. The energy of the incident photons is $10 \, eV$ and only $10\%$ of the incident photons produce photoelectrons. The number of emitted photoelectrons per second and their maximum kinetic energy, respectively, will be: $[1 \, eV = 1.6 \times 10^{-19} \, J]$

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

$A$ 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.