$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. 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

The variation of stopping potential $(V_0)$ as a function of the frequency $\nu \ (\times 10^{14} \ Hz)$ of the incident light for a metal is shown in the figure. The work function of the surface is $........... \ eV$.

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \; \mathring{A}$ is $0.42 \; V$. If the threshold frequency is $x \times 10^{13} \; s^{-1}$,where $x$ is ... (nearest integer).
(Given: speed of light $= 3 \times 10^{8} \; m/s$,Planck's constant $= 6.63 \times 10^{-34} \; J \cdot s$)

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

According to Einstein's photoelectric equation,the plot of the maximum kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope: