In photoelectric effect experiment, the stopping potential for a given metal is $\mathrm{V}_0$ (in volt) when radiation of wavelength $\lambda_0$ is used. If radiation of wavelength $5 \lambda_0$ is used with the same metal, then the stopping potential is (in volt) ( $\mathrm{h}=$ Planck's constant, $\mathrm{c}=$ velocity of light, $\mathrm{e}=$ electronic charge)

Two identical photocathodes receive light of frequencies ' $f 1$ ' and ' $f_2$ '. The velocity of the photoelectrons of mass ' $m$ ' emitted are respectively ' $\mathrm{v}_1$ ' and ' $\mathrm{v}_2$ '. Then the correct relation is ( $\mathrm{h}=$ Planck's constant)

A photon and an electron have equal energy ' $E$ '. The ratio of wavelength of photon to wavelength of electron is proportional to

The energy of a photon is equal to the kinetic energy of proton. If $\lambda_1$ is the de-Broglie wavelength of a proton, $\lambda_2$ is the wavelength associated with the photon and if E is the energy of photon then $\lambda_2: \lambda_1$ is