If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

When radiation of wavelength $\lambda$ is incident on a photocell,the maximum velocity of the photoelectrons is $v$. What will be the maximum velocity of the photoelectrons when radiation of wavelength $3\lambda/4$ is incident on the photocell?

In the case of the photoelectric effect, on increasing the frequency of incident light: