By the photoelectric effect,Einstein proved:

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

$A$ photoelectric cell is connected to a source of variable potential difference. The photoelectric current is plotted against the applied potential difference. The graph with the broken line represents the current for a given frequency and intensity of the incident radiation. If the frequency is increased and the intensity is reduced, which curve now represents the situation?

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to:

The frequency and work function of an incident photon are $\nu$ and $\phi_0$. If $\nu_0$ is the threshold frequency,then the necessary condition for the emission of photoelectrons is:

Sodium and copper have work functions $2.3 \ eV$ and $4.5 \ eV$ respectively. Then the ratio of their threshold wavelengths is nearest to