The maximum kinetic energy of the emitted photoelectrons from a photosensitive material of work function $\phi$,when light of frequency $\nu$ is incident on it,is $E$. If the frequency of the incident light is $3\nu$,the maximum kinetic energy of the emitted photoelectrons is:

When a photon of energy $hv$ is incident on an aluminum plate with a work function $E_0$,the maximum kinetic energy of the emitted photoelectrons is $K$. If the frequency of the incident radiation is doubled,the maximum kinetic energy of the emitted photoelectrons will be .......

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

Radiation of monochromatic waves with a wavelength of $400 \ nm$ is incident on the surfaces of $Zn$,$Fe$,and $Ni$ metals,which have work functions of $3.4 \ eV$,$4.8 \ eV$,and $5.9 \ eV$ respectively. (Take $hc = 1242 \ eV \ nm$)
$(a)$ The maximum $KE$ of photoelectrons emitted from any metal surface is $0.3 \ eV$.
$(b)$ No photoelectrons are emitted from the surface of $Ni$.
$(c)$ If the frequency of the radiation source is doubled,the $KE$ of the photoelectrons also doubles.
$(d)$ If the wavelength of the incident radiation is less than $200 \ nm$,photoelectrons will be emitted from the surfaces of all three metals.
The correct statements are:

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $(\lambda)$ of incident light and the corresponding stopping potential $(V_0)$ are given below:

$\lambda (\mu m)$	$V_0$ (Volt)
$0.3$	$2.0$
$0.4$	$1.0$
$0.5$	$0.4$

Given that $c = 3 \times 10^8 \ m \ s^{-1}$ and $e = 1.6 \times 10^{-19} \ C$, Planck's constant (in units of $J \ s$) found from such an experiment is:

If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then: