$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)

Photons of energy $10 eV$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15} Hz$. The kinetic energy in $eV$ of the photoelectrons emitted is [Planck's constant $h = 6.63 \times 10^{-34} Js$]. (in $eV$)

Electrons are emitted with kinetic energy $T$ from a metal plate by an irradiation of light of intensity $J$ and frequency $v$. Then, which of the following will be true?

Photoelectric effect can be explained by

$A$ 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).

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is