The work function of a photosensitive metal surface is $1.1 \ eV$. Two light beams of energies $1.5 \ eV$ and $2 \ eV$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

Why maximum kinetic energy of a photoelectron cannot be negative?

The stopping potential for a source of wavelength $4000\, \mathring{A}$,when kept at a distance of $10\, \text{cm}$,is $1.5\, \text{V}$. If now the distance of the source is increased to $20\, \text{cm}$,then the stopping potential will be ............... $\text{V}$.

$A$ photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_c$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de-Broglie wavelength is given by ($m$: mass of the electron,$h$: Planck's constant,and $c$: speed of light).

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is: