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:

Light of wavelength $\lambda$ is incident on the surface of a metal having work function $\phi$, causing the emission of electrons. What is the maximum velocity of the emitted electrons? (Given: $c = \text{velocity of light}$, $h = \text{Planck's constant}$, $m = \text{mass of electron}$)

The work function of a metal is $1.6 \times 10^{-19} \, J$. When the metal surface is illuminated by light of wavelength $6400 \, \mathring{A}$,the maximum kinetic energy of the emitted photo-electrons will be. (Planck's constant $h = 6.4 \times 10^{-34} \, Js$)

The work function of a metal is $6.825 \ eV$. Its threshold wavelength is approximately: (Given $c = 3 \times 10^8 \ m/s$)

The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is