When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $3\lambda$,the stopping potential is $\frac{V}{6}$. The threshold wavelength for the surface is:

$A$ cobalt $(Co)$ plate is placed at a distance of $1 \,m$ from a point source of power $1 \,W$. Assume a circular area of the plate of radius $r = 1 \,Å$ is exposed to the radiation and ejects photoelectrons. The light energy is considered to be spread uniformly and the work function of cobalt is $5 \,eV$. The minimum time the target should be exposed to the light source to eject a photoelectron (assuming no reflection losses) is: (in $\,s$)

Yellow light of wavelength $557 \ nm$ is incident on a cesium surface. When the cathode-anode voltage drops below $0.25 \ V$,no photoelectrons flow in the circuit. What is the threshold wavelength of the cesium surface in $nm$?

The threshold wavelength for the photoelectric effect of a metal is $6500 \mathring{A}$. The work function of the metal is approximately .......... $eV$.

Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.

For the photoelectric effect,the maximum kinetic energy $(E_{k})$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in the figure. The slope of the graph gives: