Light of frequency $4\nu_0$ is incident on a metal surface with threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

The study of photoelectric effect is useful in understanding

The maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident radiation is $\lambda$. If the wavelength of the incident radiation is reduced to $\frac{\lambda}{3}$,the maximum kinetic energy becomes $4E$. The work function of the metal is:

$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

When a photon with energy $8\ eV$ is incident on a metal surface having a threshold frequency of $1.6 \times 10^{15}\ Hz$,the maximum kinetic energy of the emitted photoelectrons is ............ $eV$. (Given: $h = 6.6 \times 10^{-34}\ J\cdot s$,$1\ eV = 1.6 \times 10^{-19}\ J$)

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 $\text{eV}$.