An electron in a Hydrogen atom jumps from the second Bohr orbit to the ground state,and the difference between the energies of the two states is radiated in the form of a photon. This photon strikes a material. If the work function of the material is $4.2 \ eV$,then the stopping potential is (Energy of electron in $n$-th orbit $= -\frac{13.6}{n^2} \ eV$). (in $V$)

In an experiment to study the photoelectric effect,the observed variation of stopping potential $(V_0)$ with the frequency $(
u)$ of incident radiation is as shown in the figure. The slope and $y$-intercept are respectively

$A$ photon of energy $4 eV$ imparts all its energy to an electron that leaves a metal surface with $1.1 eV$ of kinetic energy. The work function of the metal is (in $eV$)

$A$ 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 threshold wavelength of a metal is $400 \ nm$. The maximum kinetic energy of the emitted photoelectrons is $1.5 \ eV$. Find the wavelength of the incident photon in $\mathring{A}$.

When light of frequency $v$ is incident on a metal,photoelectrons are emitted. This is because...