The work function of a material is $4.0 \ eV$. The maximum wavelength of light that can cause the emission of photoelectrons from the material is ............ $nm$.

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 \ eV$ and $E_{2} = 2.5 \ eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal in $eV$ is:

When a light of wavelength '$\lambda$' falls on the emitter of a photosensitive surface, the maximum speed of emitted photoelectrons is '$V$'. If the incident wavelength is changed to '$2\lambda/3$',the maximum speed of emitted photoelectrons will be

Photons of energy $2.4 \text{ eV}$ and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is (in $\text{ eV}$)

$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

Find the correct statement$(s)$ about the photoelectric effect.