The metallic surface is illuminated with monochromatic light of wavelength $\lambda$ and the stopping potential for the photoelectric current is $5V_0$. When the same metallic surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. What is the threshold wavelength for the surface?

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

$A$ photon of energy $E$ ejects photoelectrons from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field with induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$,then the radius is given by

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

Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.