$A$ photon of energy $6 \ eV$ is incident on a metal surface. The work function of the metal is $2 \ eV$. The minimum reverse potential required to stop the emission of electrons is ........ $V$.

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

$A$ light source is placed at a distance of $20 \ cm$ from a photocell,and the measured stopping potential is $0.6 \ V$. If the distance of the source is changed to $40 \ cm$,what will be the stopping potential in $V$?

The work function of a metal is $2.3 \ eV$ and the wavelength of the incident photon is $4.84 \times 10^{-7} \ m$. What is the maximum kinetic energy of the photoelectrons in $eV$?

When radiation of wavelength $\lambda$ is incident on a metal,the stopping potential of photoelectrons is $4.8 \ V$. When radiation of wavelength $2\lambda$ is incident on the same metal,the stopping potential is $1.6 \ V$. What is the threshold wavelength of the metal?

When a photosensitive metal surface is illuminated with radiation of wavelength $\lambda_1$,the stopping potential is $V_1$. If the same surface is illuminated with radiation of wavelength $3\lambda_1$,the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is: