Ultraviolet radiations of $6.2 \, eV$ fall on an aluminium surface (work function $4.2 \, eV$). The kinetic energy in joules of the fastest electron emitted is approximately:

If the wavelength of incident light is decreased from $4000 \ \mathring{A}$ to $3600 \ \mathring{A}$,then the change in stopping potential will be ............. $V$.

When light is incident on a surface,photoelectrons are emitted. For these photoelectrons:

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\lambda/2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is three times that in the first case,the work function of the surface is

The photoelectric cut-off voltage in a certain experiment is $1.5 \ V$. What is the maximum kinetic energy of the emitted photoelectrons?

$A$ 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.