Ultraviolet radiations of $6.2 \, eV$ fall on an aluminium surface (work function $4.2 \, eV$). The kinetic energy in joules of the fastest electron emitted is approximately:

The threshold wavelength for the photoelectric effect on sodium is $5000\;\mathring{A}$. Its work function is:

Consider the following statements regarding the photoelectric effect experiment:
$(I)$ Photoelectrons are emitted as soon as the metal is exposed to light.
$(II)$ There is a minimum frequency below which no photocurrent is observed.
$(III)$ The stopping potential is proportional to the frequency of light.
$(IV)$ The photocurrent varies linearly with the intensity of the light.
Which of the above statements indicate that light consists of quanta (photons) with energy proportional to frequency?

Why maximum kinetic energy of a photoelectron cannot be negative?

The maximum velocity of electrons emitted from a metal surface is $V$,when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)