The work function of caesium is $2.14\, eV$. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of $0.60\, V$. (Result in $nm$)

When a photon of energy $3.8 \,eV$ falls on a metallic surface of work function $2.8 \,eV$,then the kinetic energy of the emitted electrons is .......... $eV$.

When a photosensitive material is illuminated by photons of energy $3.1 \ eV$,the stopping potential of the photoelectrons is $1.7 \ V$. When the same photosensitive material is illuminated by photons of energy $2.5 \ eV$,the stopping potential of the photoelectrons is: (in $V$)

The work functions of three metals $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. They are in decreasing order $(W_A > W_B > W_C)$. The correct graph between the maximum kinetic energy $E_k$ of the emitted electron and the frequency $v$ of the incident radiation is:

The slope of the graph between the frequency of incident light and the stopping potential for a given surface is:

The threshold frequency for a certain metal is $v_0$. When radiation of frequency $2v_0$ is incident on this metal surface,the maximum velocity of the emitted photoelectrons is $2 \times 10^6 \ m/s$. If radiation of frequency $3v_0$ is incident on the same metal surface,the maximum velocity of the emitted photoelectrons is . . . . . .