The photoelectric threshold wavelength for a certain metal surface is $3600 \mathring A$. If the metal surface is irradiated by a wavelength of $1100 \mathring A$,the kinetic energy of the emitted photoelectrons is (in $\text{ eV}$)

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 kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $k : 1$, then what is the threshold frequency of the metallic surface?

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be:

When radiation of wavelength $\lambda$ is incident on a photoelectric cell,the maximum velocity of the emitted photoelectrons is $\upsilon$. If the wavelength of the incident radiation is changed to $3\lambda / 4$,the maximum velocity of the emitted photoelectrons will be ..........

When radiation of frequency $8 \times 10^{15} \ Hz$ is incident on a metal surface with a work function of $6.125 \ eV$,what is the kinetic energy of the emitted photoelectrons in $eV$?

For the photoelectric effect,the maximum kinetic energy $(E_{k})$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in the figure. The slope of the graph gives: