$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

The work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)

The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work functions $\phi_A = 9 \, eV$ and $\phi_B = 4.5 \, eV$ in $nm$ is: (Given,$hc = 1242 \, eV \, nm$)

If the work function of a metal is $3 \ eV$,then the threshold wavelength will be ............ $\mathring{A}$.

$A$ monochromatic point source of light is placed at a distance $d$ from a metal surface. Photoelectrons are ejected at a rate $n$ per second,and with maximum kinetic energy $E$. If the source is brought nearer to a distance $d / 2$,the rate and the maximum kinetic energy per photoelectron become nearly

Which of the following graphs represents the variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident light for a photoelectric cell?