$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having work function $5.01 \, eV$,when ultraviolet light of $200 \, nm$ falls on it,must be ............... $V$.

Radiations of two photons having energies twice and five times the work function of a metal are incident successively on a metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be

On a photosensitive material,when the frequency of incident radiation is increased by $30 \%$,the kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.9 \ eV$. The work function of the surface is (in $eV$)

$K_1$ and $K_2$ are the maximum kinetic energies of photoelectrons emitted from a surface of a given material for light of wavelengths $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$, then the work function of the material is given by:

In a photoelectric emission experiment,the stopping potential for a given metal is $V$ volt,when radiation of wavelength $\lambda$ is used. If radiation of wavelength $2 \lambda$ is used with the same metal,then the stopping potential (in volt) will be. [Given: $c = \text{velocity of light}$,$e = \text{charge on electron}$,$h = \text{Planck's constant}$]