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

$A$ ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely $1$, $2$ and $3$. The threshold wavelengths of these photoelectric cells are $\lambda_1$, $\lambda_2$ and $\lambda_3$, respectively, and the magnitudes of the stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and the threshold wavelengths is $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 >> \lambda$. The correct option is:

The graph of stopping potential $V_s$ against frequency $\nu$ of incident radiation is plotted for two different metals $X$ and $Y$ as shown in the graph. If $\phi_x$ and $\phi_y$ are the work functions of $X$ and $Y$,respectively,then:

Photons with energy $5\, eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photoelectrons is $2\, eV.$ When photons of energy $6\, eV$ are incident on $C,$ no photoelectrons will reach the anode $A,$ if the stopping potential of $A$ relative to $C$ is ............ $V$.

$A$ monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function $\phi = 2.4 \text{ eV}$. These photoelectrons are made to collide with hydrogen atoms in the ground state. The maximum value of $\lambda$ for which the hydrogen atom may be ionized is [take $hc = 1240 \text{ eV-nm}$]. (in $\text{ nm}$)

When light of frequency $v$ is incident on a metal,photoelectrons are emitted. This is because...