Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about:

The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively is

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is: