CBSE Class 12 Physics Dual Nature Of Matter Frequently Asked Questions

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UNIT VII
DUAL NATURE OF MATTER

2 MARKS

1. When a monochromatic yellow coloured light beam is incident on a given photosensitive surface, photoelectrons are not ejected, while the same surface gives photoelectrons when exposed to green coloured monochromatic beam. What will happen if the surface is exposed to: (i) red coloured, monochromatic beam of light? Justify your answer. [2 Marks]

Answer:
1. No photoelectrons will be ejected when the surface is exposed to red light.
2. Green light has a higher frequency and shorter wavelength than yellow light, allowing it to exceed the threshold frequency. Red light has a lower frequency and longer wavelength than yellow light.
3. Since yellow light cannot eject photoelectrons, red light (having even lower frequency) will definitely not cause photoelectric emission.

Teacher's Note:
a) Photoelectric emission depends strictly on the frequency of incident radiation, not its intensity.
b) Order of frequencies: Red < Yellow < Green.

2. What is meant by work function of a metal? How does the value of work function influence the kinetic energy of electrons liberated during photoelectric emission? [2 Marks]

Answer:
1. Work function is the minimum amount of energy required to liberate an electron from the metal surface.
2. According to Einstein's photoelectric equation, maximum kinetic energy is given by K_max = hν − \(\phi_0\).
3. A lower work function results in a higher kinetic energy for the emitted electrons for a given frequency of incident radiation.

Teacher's Note:
a) Denoted by \(\phi_0\), it is a characteristic property of each metal.
b) Metals with low work functions require lower energy photons to eject electrons.

3. Define the terms: (i) work function, (ii) threshold frequency & (iii) stopping potential with reference of photoelectric effect. [2 Marks]

Answer:
1. Work function: The minimum energy of a photon required to eject an electron from the metal surface.
2. Threshold frequency: The minimum frequency of incident radiation below which photoelectric emission is completely absent.
3. Stopping potential: The minimum negative potential applied to the anode that completely stops the photoelectric current.

Teacher's Note:
a) These three terms form the foundational basis of the experimental study of the photoelectric effect.
b) Stopping potential depends directly on the frequency of incident light.

4. The work function of lithium is 2.3 eV. What does it mean? What is the relation between the work function 'ωo' and threshold wavelength 'λo' of a metal? [2 Marks]

Answer:
1. It means an energy of 2.3 eV is required to liberate an electron from the surface of lithium.
2. The relation between work function \(\omega_0\) and threshold wavelength \(\lambda_0\) is given by \(\omega_0\) = \(\frac{hc}{\lambda_0}\).

Teacher's Note:
a) 1 eV = 1.6 × 10^-19 Joules.
b) Work function and threshold wavelength are inversely proportional to each other.

5. Red light, however bright, cannot cause emission of electrons from a clean zinc surface. But, even weak ultraviolet radiations can do so. Why? [2 Marks]

Answer:
1. The frequency of red light is less than the threshold frequency of zinc, whereas ultraviolet radiation has a frequency much greater than the threshold frequency.
2. Photoelectric emission is a frequency-dependent phenomenon, not an intensity-dependent one.

Teacher's Note:
a) Brightness only affects the number of photons, not their individual energy.
b) A single ultraviolet photon has enough energy to eject an electron, unlike a red light photon.

6. An electron and a proton have same kinetic energy. Which of the two has a greater wavelength? Explain. [2 Marks]

Answer:
1. De-Broglie wavelength is given by \(\lambda\) = \(\frac{h}{\sqrt{2mE}}\).
2. Since kinetic energy E and Planck's constant h are the same for both, wavelength is inversely proportional to the square root of mass (\(\lambda \propto \frac{1}{\sqrt{m}}\)).
3. Since the mass of an electron is much smaller than the mass of a proton, the electron has a greater wavelength.

Teacher's Note:
a) Mass of proton is approximately 1836 times the mass of an electron.
b) Lighter particles possess larger de-Broglie wavelengths for identical kinetic energies.

7. Define the term threshold frequency & work function in relation to photoelectric effect. [2 Marks]

Answer:
1. Threshold frequency: The minimum frequency of incident radiation below which no photoelectrons are emitted from the metal surface.
2. Work function: The minimum energy needed by an electron to escape from the metal surface.

Teacher's Note:
a) Both concepts explain why low-frequency radiation fails to show the photoelectric effect.
b) Mathematically linked via the equation \(\phi_0\) = h\(\nu_0\).

8. An electron and a proton are moving in the same direction and possess same kinetic energy. Find the ratio of de-Broglie wavelengths associated with these particles. [2 Marks]

Answer:
1. Formula for de-Broglie wavelength is \(\lambda\) = \(\frac{h}{\sqrt{2mE}}\).
2. Ratio of wavelengths for electron and proton: \(\frac{\lambda_e}{\lambda_p}\) = \(\sqrt{\frac{m_p}{m_e}}\).
3. Since \(m_p\) > \(m_e\), the ratio \(\frac{\lambda_e}{\lambda_p}\) = \(\sqrt{\frac{m_p}{m_e}}\): 1.

Teacher's Note:
a) Substitute mass values to find the exact numerical ratio if required.
b) Ensure proper square root simplification in final steps.

9. In the photoelectric effect experiment, the graph between the stopping potential V and frequency \(\nu\) of the incident radiation on two different metal plates P and Q are shown in the figure. (i) Which of the two metal plates P and Q has greater value of work function? (ii) What does the slope of the line depict? [2 Marks]

Answer:
1. (i) Metal plate Q has a greater value of work function because its threshold frequency is higher (\(\nu_{0Q} > \nu_{0P}\)), and work function is given by \(W = h\nu_0\).
2. (ii) The slope of the stopping potential versus frequency graph represents the ratio of Planck constant to electronic charge, which is \(\frac{h}{e}\).

Teacher's Note:
a) The Einstein photoelectric equation is \(eV = h\nu - W\), which rearranges to \(V = \left(\frac{h}{e}\right)\nu - \frac{W}{e}\).
b) The intercept on the frequency axis gives the threshold frequency, while the intercept on the potential axis relates to the work function.

3 MARKS

10. What is photoelectric effect? Write Einstein's photoelectric equation and use it to explain: (i) independence of maximum energy of emitted photoelectrons from the intensity of incident light. (ii) Existence of a threshold frequency for the emission of photoelectrons. [3 Marks]

Answer:
1. Photoelectric effect is the phenomenon of emission of electrons from a metal surface when light of suitable frequency falls on it.
2. Einstein's photoelectric equation is \(K_{max} = h\nu - \phi_0\), where \(K_{max}\) is the maximum kinetic energy, \(h\nu\) is the energy of the incident photon, and \(\phi_0\) is the work function.
3. (i) Increasing the intensity of incident light increases the number of photons striking per second, thereby increasing the number of emitted electrons, but it does not change the energy of individual photons, so maximum kinetic energy remains independent of intensity.
4. (ii) Emission of photoelectrons occurs only when photon energy exceeds work function (\(h\nu > \phi_0\)), which defines the threshold frequency \(\nu_0 = \frac{\phi_0}{h}\); frequencies below this cannot eject electrons.

Teacher's Note:
a) The wave theory failed to explain the instantaneous nature and frequency dependence of the photoelectric effect.
b) Einstein explained it successfully by treating light as a stream of energy packets called photons.
c) Always state the conservation of energy principle behind Einstein's equation.

11. Draw the variation of maximum kinetic energy of emitted electrons with frequency of the incident radiation on a photosensitive surface. On the graph drawn, what do the following indicate: (i) slope of the graph & (ii) intercept on the energy axis. [3 Marks]

Answer:
1. The graph is a straight line starting from the threshold frequency \(\nu_0\) on the frequency axis and rising linearly with frequency.
2. (i) The slope of the graph represents the Planck constant (\(h\)).
3. (ii) The negative intercept on the energy axis represents the work function (\(\phi_0\)) of the metal surface.

Teacher's Note:
a) The equation of the straight line is \(K = h\nu - \phi_0\).
b) The slope remains constant for all metal surfaces, while the threshold frequency changes according to the work function.
c) Practice drawing axes clearly with proper labels for full credit in board exams.

12. Obtain Einstein’s photoelectric equation. Explain how it enables us to understand the (i) linear dependence of the maximum kinetic energy of the emitted electrons, on the frequency of the incident radiation & (ii) existence of a threshold frequency for a given photo emitter. [3 Marks]

Answer:
1. When a photon of energy \(h\nu\) falls on a metal surface, part of its energy is used to bind the electron free from the surface (work function \(\phi_0\)) and the rest appears as maximum kinetic energy \(K_{max}\) of the emitted photoelectron.
2. By conservation of energy, \(h\nu = \phi_0 + K_{max}\), which gives Einstein's photoelectric equation as \(K_{max} = h\nu - \phi_0\).
3. (i) Since \(h\) and \(\phi_0\) are constants for a given metal, \(K_{max}\) varies linearly with the frequency \(\nu\) of incident radiation.
4. (ii) For emission to take place, \(K_{max}\) must be greater than or equal to zero, so \(h\nu \ge \phi_0\) or \(\nu \ge \frac{\phi_0}{h}\), establishing the existence of threshold frequency \(\nu_0\).

Teacher's Note:
a) Mention energy conservation explicitly when deriving the equation.
b) Clearly distinguish between work function and threshold frequency in the explanation.
c) This derivation is a frequent repeating question in board examinations.

13. Given below is the graph between frequency (\(\nu\)) of the incident light and maximum kinetic energy (E) of emitted photoelectrons. Find the values of: (i) threshold frequency and (ii) work function from the graph. [3 Marks]

Answer:
1. (i) The threshold frequency is obtained from the intercept of the graph lines on the frequency axis (\(\nu\) axis) where kinetic energy is zero.
2. (ii) The work function for each metal is obtained by multiplying the Planck constant with its respective threshold frequency, given by \(\phi_0 = h\nu_0\), or by finding the magnitude of the negative intercept on the energy axis.
3. From the given graph, Cesium has a lower threshold frequency compared to Calcium, hence Cesium has a smaller work function.

Teacher's Note:
a) The intercept on the energy axis gives the work function directly if extrapolated.
b) Different metals have different threshold frequencies due to different binding energies of electrons.
c) Comparative graph reading questions test conceptual clarity of photoelectric parameters.

14. Sketch a graph between frequency of incident radiations and stopping potential for a given photosensitive materials. What information can be obtained from the value of intercept on the potential axis? A source of light of frequency greater that the threshold frequency is replaced at a distance of 1 m from the cathode of a photo cell. The stopping potential is found to be V. If the distance of the light source from the cathode is reduced, explain giving reason, what change you will observe in the (i) photoelectric current & (ii) stopping potential. [3 Marks]

₀ -W/e

Answer:
1. The intercept on the negative potential axis gives the value of \(\frac{W}{e}\), where W is the work function of the photosensitive material and e is the electronic charge.
2. (i) When the distance of the light source from the cathode is reduced, the intensity of incident light increases because intensity is inversely proportional to the square of the distance (\(I \propto \frac{1}{r^2}\)). Since photoelectric current is directly proportional to intensity, the photoelectric current increases.
3. (ii) The stopping potential remains unchanged at V because stopping potential depends only on the frequency of the incident radiation and the work function of the cathode material, neither of which changes when the distance is reduced.

Teacher's Note:
a) Einstein's photoelectric equation is given by \(eV_s = h\nu - W\), which forms the basis of the straight line graph.
b) Students often incorrectly assume that changing the distance or intensity affects the stopping potential, but kinetic energy of emitted photoelectrons is independent of intensity.
c) The slope of the frequency versus stopping potential graph represents the universal constant \(\frac{h}{e}\).

15. Explain the laws of photoelectric emission on the basis of Einstein’s photoelectric equation. Write one feature of the photoelectric effect which cannot be explained on the basis of wave theory of light. [3 Marks]

Answer:
1. Einstein's photoelectric equation is \(K_{max} = h\nu - \phi_0\), where \(h\nu\) is the energy of the incident photon and \(\phi_0\) is the work function.
2. Law 1 (Threshold Frequency): Emission occurs only if \(h\nu \gt \phi_0\). If frequency is less than threshold frequency (\(\nu_0\)), kinetic energy becomes negative, which is impossible, explaining why below \(\nu_0\) no emission takes place regardless of intensity.
3. Law 2 (Intensity and Current): Photoelectric current is directly proportional to the intensity of incident radiation because higher intensity means more photons striking per second, releasing more electrons.
4. Law 3 (Maximum Kinetic Energy): Maximum kinetic energy depends linearly on the frequency of incident radiation and is completely independent of its intensity, as a single photon interacts with a single electron.
5. Feature unexplained by wave theory: Wave theory cannot explain the instantaneous emission of photoelectrons without any time lag, regardless of how low the intensity of the incident radiation is.

Teacher's Note:
a) Einstein successfully used Planck's quantum theory to explain photoelectric emission using energy conservation.
b) A common mistake is stating that intensity affects electron kinetic energy, whereas wave theory falsely predicted that it would.
c) Mentioning the absence of time lag is crucial when asked about failures of wave theory in the photoelectric effect.

16. Draw graphs showing the variation of photoelectric current with anode potential of a photocell for (i) the same frequency but different intensities \(I_1 \gt I_2 \gt I_3\) of incident radiation, & (ii) the same intensity but different frequencies \(\nu_1 \gt \nu_2 \gt \nu_3\) of incident radiation. Explain why the saturation current is independent of the anode potential? [3 Marks]

₁ I₂ I₃ -V_s

Answer:
1. The saturation current is independent of the anode potential because once the positive potential of the anode is sufficiently high, all the photoelectrons emitted by the cathode are successfully collected by the anode.
2. Increasing the anode potential further cannot collect any more electrons since the total number of electrons emitted per second by the cathode depends solely on the intensity of the incident radiation.
3. For case (i), curves show different saturation currents with the same stopping potential because intensity changes while frequency remains constant.
4. For case (ii), curves show the same saturation current with different stopping potentials because frequency changes while intensity remains constant.

Teacher's Note:
a) Saturation current represents the maximum possible photoelectric current for a given intensity.
b) Students often confuse the effects of frequency and intensity on saturation and stopping potentials.
c) Clear labeling of axes and individual curves is essential for full credit in graph-based questions.