Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

When a photosensitive surface is irradiated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is:

$A$ 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 nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

The retarding potential necessary to stop the emission of photoelectrons,when a target material of work function $1.24 eV$ is irradiated with light of wavelength $4.36 \times 10^{-7} m$ is (in $eV$)