NEET Modern Physics MCQs & PYQs

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.

From Einstein's photoelectric equation, the initial maximum kinetic energy is $K = h\nu - E_0$ . When the frequency is doubled, the new incident energy is $2h\nu$ . The new maximum kinetic energy is $K' = 2h\nu - E_0$ . This can be rewritten as $K' = h\nu + (h\nu - E_0) = h\nu + K$ .

In a discharge tube, a high potential difference accelerates free electrons (cathode rays) to high speeds. The ionization of the enclosed gas is primarily caused by the energetic collisions between these rapidly moving negative electrons and the neutral gas atoms or molecules.

A photocell works on the principle of the photoelectric effect, converting light energy into electrical energy. Specifically, an increase in the intensity of illumination increases the number of emitted photoelectrons per second, converting the change in intensity into a proportional change in photoelectric current.

Initially, $\frac{1}{2}mv_1^2 = E_1 - W = 2h\nu_0 - h\nu_0 = h\nu_0$ . Finally, $\frac{1}{2}mv_2^2 = E_2 - W = 5h\nu_0 - h\nu_0 = 4h\nu_0$ . Taking the ratio gives $(\frac{v_2}{v_1})^2 = 4$ , so $v_2 = 2v_1$ . Substituting the given velocity, $v_2 = 2 \times (4 \times 10^6) = 8 \times 10^6 m/s$ .

The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{12400}{4100} eV \approx 3.02 eV$ . For photoelectric emission to occur, the incident energy must be greater than the work function ( $E > W$ ). Since $3.02 eV$ is greater than the work functions of A ( $1.92 eV$ ) and B ( $2.0 eV$ ) but less than C ( $5 eV$ ), only A and B will emit photoelectrons.

Intensity of illumination is inversely proportional to the square of distance ( $I \propto \frac{1}{d^2}$ ). When the distance is shifted from 1 m to 2 m, it doubles, so the intensity becomes one-fourth ( $1/4$ ) of its initial value. Since the number of emitted electrons is directly proportional to intensity, it is reduced to a quarter of the initial number.