The figure shows the variation of photocurrent with anode potential for four different radiations. Let $f_a$,$f_b$,$f_c$ and $f_d$ be the frequencies for the curves $a$,$b$,$c$ and $d$ respectively.

$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

Electrons ejected from the surface of a metal,when light of a certain frequency is incident on it,are stopped fully by a retarding potential of $3 \ V$. The photoelectric effect on this metallic surface begins at a frequency of $6 \times 10^{14} \ s^{-1}$. The frequency of the incident light in $s^{-1}$ is: [Planck's constant $= 6.4 \times 10^{-34} \ J \cdot s$,charge on the electron $= 1.6 \times 10^{-19} \ C$]

When light of frequency $v$ is incident on two metallic plates $A$ and $B$,photoelectrons are emitted. If the work function of $A$ is greater than that of $B$,which of the following curves correctly represents the relationship between the stopping potential $V$ and the incident frequency $v$?

$K_1$ and $K_2$ are the maximum kinetic energies of photoelectrons emitted from a surface of a given material for light of wavelengths $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$, then the work function of the material is given by: