In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

$A$ photoelectric cell is connected to a source of variable potential difference and the resulting photoelectric current $(\mu A)$ is plotted against the applied potential difference $(V)$. The graph with the broken line represents one situation for a given frequency and intensity of the incident radiation. If the frequency is increased and intensity is reduced,which of the following graphs of unbroken line represents the new situation?

Photons of frequencies equal to the frequencies of $H_\beta$ and $H_{\infty}$ lines of hydrogen are incident on a photosensitive plate,whose threshold frequency is equal to the frequency of the $H_\alpha$ line of hydrogen. The ratio of the maximum kinetic energies of the emitted electrons is

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

Light of wavelength $5000 \; \mathring A$ falls on a sensitive plate with photoelectric work function of $1.9 \; eV$. The kinetic energy of the photoelectron emitted will be ............ $eV$.