The threshold frequency of cesium is $5.16 \times 10^{14} \ Hz$. Then its work function is . . . . . . $eV$.

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

Light of wavelength '$\lambda$' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity, then the stopping potential:

Light of frequency $4v_0$ is incident on a metal surface with a threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is:

The work function of caesium is $2.14 \ eV$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$.

Light is incident on a metallic plate having a work function of $ 110 \times 10^{-20} \ J $. If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is . . . . . . $ rad/s $. $( h = 6.63 \times 10^{-34} \ J \cdot s )$