The surface of a metal is illuminated with light of $400 \ nm$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \ eV$. The work function of the metal is ............ $eV$ $(hc = 1240 \ eV \ nm)$.

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

For zero photoelectric current,the stopping potential is:

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14 \text{ eV}$ and the stopping potential is $2 \text{ V}$, what is the wavelength of the electromagnetic wave (in $\text{ nm}$)? (Given $hc = 1242 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)

Let $v$ be the frequency of the incident photon and $v_0$ be the threshold frequency. The condition for the emission of photoelectrons is:

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?