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

Einstein's photoelectric equation is $E_k = hf - \phi_0$. In this equation,$E_k$ represents ...........

When the energy of incident radiation is increased by $20 \%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5 \ eV$ to $0.8 \ eV$. The work function of the metal is (in $eV$)

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons, moving perpendicular to the field, are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$, then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

The electric field of a light wave is given as $\vec E = 10^{-3} \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat x \, N/C$. This light falls on a metal plate with a work function of $2 \, eV$. The stopping potential of the photoelectrons is ................ $V$.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\lambda /2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case,the work function of the surface of the material is
$(h =$ Planck's constant,$c =$ speed of light $)$