When monochromatic light falls on a photo-sensitive metal,an electron is emitted with maximum velocity $1.6 \times 10^6 \ m/s$. Find the stopping potential.
[charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$] (in $V$)

The electron in the hydrogen atom jumps from excited state $(n = 3)$ to its ground state $(n = 1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \, eV$,the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n = \frac{-13.6}{n^2} \, eV$) (in $, V$)

Light of frequency $\nu$ is incident on a substance of threshold frequency $\nu_0$ (where $\nu_0 < \nu$). The maximum kinetic energy of the emitted photo-electron will be:

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:

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

Photoelectrons are emitted from two similar metal plates when wavelengths $\lambda_1$ and $\lambda_2$ are incident on them $(\lambda_1 = 1.5\lambda_2)$. If the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively, then: