The wave nature of light cannot explain the photoelectric effect because,in the photoelectric effect,it is observed that:

Match the following columns:

Column-$I$	Column-$II$
$A$. Hallwachs $\&$ Lenard	$P$. Transformers
$B$. Franck-Hertz	$Q$. Microwave
$C$. Klystron valve	$R$. Quantization of energy levels
$D$. Nicola Tesla	$S$. Photoelectric effect

An electron moving with speed $v$ and a photon moving with speed $c$ have the same de Broglie wavelength. The ratio of the kinetic energy of the electron to that of the photon is -

In a photoelectric effect experiment,three different metal plates $p, q,$ and $r$ have work functions $\phi_p = 2.0 \ eV, \phi_q = 2.5 \ eV,$ and $\phi_r = 3.0 \ eV$. The incident light beams have wavelengths $\lambda_p = 550 \ nm, \lambda_q = 450 \ nm,$ and $\lambda_r = 350 \ nm$. The intensity of light is the same for each plate. The correct $I-V$ graph for the experiment is: $[hc = 1240 \ eV \ nm]$

The correct curve between the stopping potential $(V_s)$ and intensity of incident light $(I)$ is

$A$ light beam of wavelength $500 \, nm$ is incident on a metal having a work function of $1.25 \, eV$,placed in a magnetic field of intensity $B$. The electrons emitted perpendicular to the magnetic field $B$,with maximum kinetic energy,are bent into a circular arc of radius $30 \, cm$. The value of $B$ is $.... \times 10^{-7} \, T$.
Given $hc = 20 \times 10^{-26} \, J \cdot m$,mass of electron $m_e = 9 \times 10^{-31} \, kg$,and $e = 1.6 \times 10^{-19} \, C$.