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

The threshold frequency for a certain metal is $v_0$. When radiation of frequency $2v_0$ is incident on this metal surface,the maximum velocity of the emitted photoelectrons is $2 \times 10^6 \ m/s$. If radiation of frequency $3v_0$ is incident on the same metal surface,the maximum velocity of the emitted photoelectrons is . . . . . .

When the wavelength of an incident photon is decreased, then:

$A$ photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$,then the radius $r$ is given by,(in the usual notation)

$A$ material $A$ exhibits the photoelectric effect. Its work function is $2.5 \ eV$ and the threshold wavelength is $\lambda$. Another material $B$ has a work function of $5 \ eV$. Find the threshold wavelength required to produce the photoelectric effect in $B$.

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is