In a photoelectric experiment,if the wavelength of incident radiation is reduced from $6000 \ \mathring{A}$ to $4000 \ \mathring{A}$ while keeping the intensity of radiation constant,then:

The work function of a metal is $1 \; eV$. Light of wavelength $3000 \; \mathring{A}$ is incident on this metal surface. The velocity of the emitted photo-electrons will be:

Photoelectric emission is observed from a metallic surface for frequencies ${\nu _1}$ and ${\nu _2}$ of the incident light rays $({\nu _1} > {\nu _2})$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ J \ m\right]$

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photoelectrons (of mass $m$) coming out are respectively $v_1$ and $v_2$,then:

If the threshold wavelength for sodium is $5420 \mathring{A}$,then the work function of sodium is ............ $eV$.