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:

$A$ metal surface of work function $1.13 eV$ is irradiated with light of wavelength $310 nm$. The retarding potential required to stop the escape of photoelectrons is [Take $hc = 1240 eV \cdot nm$] (in $V$)

The stopping potential required to reduce the photoelectric current to zero is . . . . . .

An image of the sun is formed by a lens of focal length $30 \ cm$ on the metal surface of a photoelectric cell and a photoelectric current $I$ is produced. The lens forming the image is then replaced by another of the same diameter but of focal length $15 \ cm$. The photoelectric current in this case is

In the graph given below,if the slope is $4.12 \times 10^{-15} \, V-s$,then the value of $'h'$ should be:

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]