In a photoelectric effect experiment,the cathode metal is exposed to light of wavelength $600 \ nm$. The maximum kinetic energy of the ejected electron doubles when light of wavelength $400 \ nm$ is used. The work function of the cathode metal is approximately: [Use $h=6.63 \times 10^{-34} \ J-s, c=3 \times 10^8 \ m/s$] (in $eV$)

Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons emitted are $v_{1}$ and $v_{2}$ respectively,then:

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda / 3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case,the work function of the surface of the material is ($h=$ Planck's constant,$c=$ speed of light).

The work function of Cesium is $2.27 eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 nm$ wavelength is

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ J \ m\right]$

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