$A$ metal is exposed to light of wavelength $800 \, nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of the photoelectrons doubles when light of wavelength $500 \, nm$ is used. The work function of the metal is $... \, eV$ (Take $hc = 1230 \, eV \cdot nm$).

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

Explain: "For photoelectric emission, the frequency of incident radiation should be greater than the threshold frequency."

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

When a photon of energy $4.0 \; eV$ strikes the surface of a metal $A$,the ejected photoelectrons have maximum kinetic energy $T_{A} \; eV$ and de-Broglie wavelength $\lambda_{A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.

The work function of a metal is $2.51 eV$. Its threshold frequency is: