Light of wavelength $5000 \; \mathring A$ falls on a sensitive plate with photoelectric work function of $1.9 \; eV$. The kinetic energy of the photoelectron emitted will be ............ $eV$.

$A$ monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function $\phi = 2.4 \text{ eV}$. These photoelectrons are made to collide with hydrogen atoms in the ground state. The maximum value of $\lambda$ for which the hydrogen atom may be ionized is [take $hc = 1240 \text{ eV-nm}$]. (in $\text{ nm}$)

In an experiment aimed to disprove Einstein's photoelectric equation,how did Millikan prove it?

When photons of energy $hv$ fall on a metal plate of work function $W_0$,photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of the radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be

In the graph given below,if the slope is $4.12 \times 10^{-15} \, V-s$,then the value of $'h'$ should be:

The stopping potential $(V_s)$ as a function of the frequency $(\nu)$ of the incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graphs show that the work function of $A$ is