Light of frequency $8 \times 10^{15} \ Hz$ is incident on a substance of photoelectric work function $6.125 \ eV$. The maximum kinetic energy of the emitted photoelectrons is ........... $eV$.

Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

The photoelectric work function of a metal is $1 \ eV$. Light of wavelength $\lambda = 3000 \ \mathring{A}$ is incident on it. What will be the velocity of the emitted photoelectrons?

If light of wavelength $6600 \ \mathring A$ is incident on a metal surface with a work function of $2 \ eV$,what will be the maximum kinetic energy of the emitted photoelectrons?

In a photoelectric effect experiment,the maximum kinetic energy of emitted photoelectrons is $K_0$. If the frequency of the incident radiation is increased by a factor of $n_1$,the new maximum kinetic energy becomes $n_2K_0$. Find the work function of the metal.