The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is $25 \%$ of the velocity of light,then the ratio of the $K.E.$ of the electron to the $K.E.$ of the photon will be:

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

The spectrum obtained from an electric lamp or a red-hot heater 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 de Broglie wavelength of a particle moving with a speed of $0.8 c$ is equal to the wavelength of a photon. If $c$ is the speed of light in vacuum, the ratio of the energy of the photon to the kinetic energy of the particle is:

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