In case of photoelectric emission from a certain metal, the cutoff frequency is $\nu$. If radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant).

Einstein's photoelectric equation states that ${E_k} = h\nu - \phi$. In this equation,${E_k}$ refers to:

Light of frequency $4\nu_0$ is incident on a metal surface with a threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

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:

The work functions of three metals $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. They are in decreasing order $(W_A > W_B > W_C)$. The correct graph between the maximum kinetic energy $E_k$ of the emitted electron and the frequency $v$ of the incident radiation is:

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)