The work function of a substance is $3.0 \ eV$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately: (in $nm$)

The maximum velocity of photoelectrons emitted from a metal surface is $1.2 \times 10^6 \ m/s$. If the specific charge of the electron is $1.8 \times 10^{11} \ C/kg$,then the value of the stopping potential in volts is:

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

The stopping potential $(V_s)$ as a function of the frequency $(\nu)$ of the incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graphs show that the work function of $A$ is

The maximum velocity of electrons emitted from a metal surface is $V$,when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?