When photons of energy $hf$ fall on an aluminium plate (of work function $E_0$),photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled,the maximum kinetic energy of the ejected 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.

The work function of a metal is $1 \ eV$. If light of wavelength $3000 \ \mathring{A}$ is incident on the surface of this metal, the maximum velocity of the emitted photoelectrons is ........

When the wavelength of an incident photon is decreased, then:

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\lambda /2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case,the work function of the surface of the material is
$(h =$ Planck's constant,$c =$ speed of light $)$

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be: