Question 61:
easy18. Photoelectric emission occurs only when the incident light has more than a certain minimum: (2011 Pre)
Photoelectric emission takes place only when the frequency of the incident light is greater than a characteristic minimum frequency for the given metal, known as the threshold frequency.
Question 62:
easy31. A photosensitive metallic surface has work function, $h\nu_0$ . If photons of energy $2h\nu_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 m/s$ . When the photon energy is increased to $5h\nu_0$ , then maximum velocity of photoelectrons will be: (2005)
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, $v_2 = 2 \times (4 \times 10^6) = 8 \times 10^6 m/s$ .
Question 63:
easy36. Which of the following is not the property of cathode rays: (2002)
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.
Question 64:
easy37. Which one among the following shows particle nature of light: (2001)
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.
Question 65:
easy38. A photo-cell is illuminated by a source of light, which is placed at a distance d from the cell. If the distance become d/2, then number of electrons emitted per second will be: (2001)
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.
Question 66:
easy39. By photoelectric effect, Einstein proved: (2000)
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 $E = h\nu$ .
Question 67:
easy40. Who evaluated the mass of electron indirectly with the help of charge: (2000)
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.
Question 68:
easy41. The current conduction in a discharge tube is due to: (1999)
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.
Question 69:
easy42. Light of wavelength $3000 \mathring{A}$ in Photoelectric effect gives electron of max. K.E. 0.5 eV. If wavelength change to $2000 \mathring{A}$ then max. K.E. of emitted electrons will be: (1999)
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.
Question 70:
easy43. If the light of wavelength $\lambda$ is incident on metal surface, the ejected fastest electron has speed v. If the wavelength is changed to $\frac{3\lambda}{4}$ the speed of the fastest emitted electron will be: (1998)
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 $v' > \sqrt{\frac{4}{3}}v$ .