Two identical metal plates show the photoelectric effect. Light of wavelength ${\lambda _A}$ falls on plate $A$ and light of wavelength ${\lambda _B}$ falls on plate $B$,where ${\lambda _A} = 2{\lambda _B}$. The maximum kinetic energy is:

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14 \text{ eV}$ and the stopping potential is $2 \text{ V}$, what is the wavelength of the electromagnetic wave (in $\text{ nm}$)? (Given $hc = 1242 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)

$A$ 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 ........ $V$.

The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. $A$ light source is turned on and a saturation photocurrent is recorded. An electric field is then switched on which has a vertically downward direction.

$A$ stream of photons having energy $3 \,eV$ each impinges on a potassium surface. The work function of potassium is $2.3 \,eV$. The emerging photo-electrons are slowed down by a copper plate placed $5 \,mm$ away. If the potential difference between the two metal plates is $1 \,V$,the maximum distance the electrons can move away from the potassium surface before being turned back is .......... $mm$.