The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

Monochromatic light of wavelength $\lambda = 4770 \ \mathring{A}$ is incident separately on the surface of four different metals $A, B, C$ and $D$. The work functions of $A, B, C$ and $D$ are $4.2 \ \text{eV}, 3.7 \ \text{eV}, 3.2 \ \text{eV}$ and $2.3 \ \text{eV}$,respectively. From which of these metals will electrons be emitted?

An isolated lead ball is charged upon continuous irradiation by $EM$ radiation of wavelength, $\lambda = 221 \,nm$. The maximum potential attained by the lead ball, if its work function is $4.14 \,eV$, is (take, $h = 6.63 \times 10^{-34} \,J \cdot s$, $c = 3 \times 10^8 \,m/s$, $e = 1.6 \times 10^{-19} \,C$): (in $\,V$)

If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

The maximum kinetic energy of emitted photoelectrons depends on

Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4v$ is incident on the metal plate,the maximum velocity of the emitted photoelectrons will be ($m=$ mass of photoelectron,$h=$ Planck's constant).