The work-function of a metal is $1 eV$. Light of wavelength $3000 \text{Å}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

The frequency and work function of an incident photon are $\nu$ and $\phi_0$. If $\nu_0$ is the threshold frequency,then the necessary condition for the emission of photoelectrons is:

When light is incident on a surface,photoelectrons are emitted. For these photoelectrons:

When a photon of energy $hv$ is incident on an aluminum plate with a work function $E_0$,the maximum kinetic energy of the emitted photoelectrons is $K$. If the frequency of the incident radiation is doubled,the maximum kinetic energy of the emitted photoelectrons will be .......

When a metallic surface is illuminated by light of wavelength $\lambda$,the stopping potential for the photoelectric current is $3 \text{ V}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $1 \text{ V}$. The threshold wavelength for this surface is:

From a metallic surface,photoelectric emission is observed for frequencies $v_1$ and $v_2$ $(v_1 > v_2)$ of the incident light. The maximum kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence,the threshold frequency of the metallic surface is