Light is incident on a metallic plate having a work function of $ 110 \times 10^{-20} \ J $. If the produced photoelectrons have zero kinetic energy,then the angular frequency of the incident light is . . . . . . $ rad/s $. $( h = 6.63 \times 10^{-34} \ J \cdot s )$

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is:

When a photosensitive surface is irradiated by light of wavelengths $\lambda_{1}$ and $\lambda_{2}$,the kinetic energies of the emitted photoelectrons are $E_{1}$ and $E_{2}$ respectively. The work function of the photosensitive surface is

In a photoelectric effect experiment, the frequency of incident radiation is increased while keeping other factors constant $(f > f_0)$. The stopping potential

In the photoelectric effect,the stopping potential $(V_0)$ versus frequency $(\nu)$ curve is plotted. ($h$ is Planck's constant and $\phi_0$ is the work function of the metal)
$(A)$ $V_0$ versus $\nu$ is linear.
$(B)$ The slope of the $V_0$ versus $\nu$ curve $= \frac{\phi_0}{h}$.
$(C)$ Planck's constant $h$ is related to the slope of the $V_0$ versus $\nu$ line.
$(D)$ The value of the electric charge of an electron is not required to determine $h$ using the $V_0$ versus $\nu$ curve.
$(E)$ The work function can be estimated without knowing the value of $h$.
Choose the correct answer from the options given below:

The stopping potential $(V_0)$ versus frequency $(\nu)$ plot of a substance is shown in the figure. The threshold wavelength is: