NEET Modern Physics MCQs & PYQs

If ultraviolet rays cannot induce the photoelectric effect, it means their frequency is below the threshold frequency. To cause emission, a radiation with a higher frequency (and higher photon energy) is needed. Among the options, only X-rays have a higher frequency and energy than ultraviolet rays.

Cathode rays are streams of fast-moving electrons, which are negatively charged particles. Because they carry an electric charge, they are readily deflected by both electric and magnetic fields. Therefore, the statement that they do not deflect in an electric field is incorrect.

Phenomena such as interference, diffraction, and polarization are successfully explained by the wave theory of light. The photoelectric effect, however, can only be explained by assuming that light consists of discrete energy packets or particles called photons, demonstrating its particle nature.

The intensity of incident light $I$ varies with distance as $I \propto \frac{1}{d^2}$ . When the distance is halved to $d/2$ , the intensity becomes $\frac{1}{(1/2)^2} = 4$ times its original value. Since the number of photoelectrons emitted per second is directly proportional to intensity, it also becomes four times.

Albert Einstein used Max Planck's quantum hypothesis to successfully explain the photoelectric effect. He proved that light interacts with matter as discrete quanta of energy (photons), where the energy of each photon is given by $E = h\nu$ .

J.J. Thomson discovered the electron and determined its specific charge (the charge-to-mass ratio, $e/m$ ). Once Robert Millikan independently measured the fundamental charge ( $e$ ), Thomson's ratio allowed for the indirect evaluation of the electron's mass.

In a gas discharge tube, the applied high voltage ionizes the gas atoms. This creates free electrons and positive ions. Both of these charged particles migrate towards opposite electrodes under the electric field, contributing to the current conduction.

Incident energy $E = \frac{hc}{\lambda}$ . Initially, $E_1 = \frac{12400}{3000} \approx 4.13 eV$ . Work function is $W = E_1 - K_1 = 4.13 - 0.5 = 3.63 eV$ . For $2000 \mathring{A}$ , new energy is $E_2 = \frac{12400}{2000} = 6.2 eV$ . The new maximum kinetic energy is $K_2 = E_2 - W = 6.2 - 3.63 = 2.57 eV$ , which is clearly greater than 0.5 eV.

Initially, $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - W$ . Finally, $\frac{1}{2}mv'^2 = \frac{hc}{3\lambda/4} - W = \frac{4hc}{3\lambda} - W$ . This can be rewritten as $\frac{1}{2}mv'^2 = \frac{4}{3}(\frac{hc}{\lambda} - W) + \frac{W}{3} = \frac{4}{3}(\frac{1}{2}mv^2) + \frac{W}{3}$ . Since $W$ is positive, $\frac{1}{2}mv'^2 > \frac{4}{3}(\frac{1}{2}mv^2)$ , meaning $v'^2 > \frac{4}{3}v^2$ or $v' > \sqrt{\frac{4}{3}}v$ .

The intensity of light $I$ is inversely proportional to the square of the distance $r$ from a point source, so $I \propto \frac{1}{r^2}$ . When the distance is doubled from $0.5 m$ to $1.0 m$ , the intensity becomes $\frac{1}{4}$ of its initial value. Since the number of photoelectrons liberated is directly proportional to intensity, it will also be reduced by a factor of 4.