The number of photons falling per second on a completely darkened plate to produce a force of $6.62 \times 10^{-5} \mathrm{~N}$ is $n$. If the wavelength of the light falling is $5 \times 10^{-7} \mathrm{~m}$, then $n=$ $\ldots \ldots \ldots \times 10^{22}$. (Take, $h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

The maximum kinetic energy of emitted photoelectrons depends on

A proton and an $\alpha$-particle are accelerated through the same potential difference $V$. The ratio of their de-Broglie wavelength is

From the following graph of photo current against collector plate potential, for two different intensities of light $I_1$ and $I_2$, one can conclude