Chapter 8 — Exercise Solutions
This chapter is scoring if you separate photoelectric effect formulas from de Broglie wavelength formulas.
How to Use This Page
- Start with the Key Concepts table to refresh definitions before attempting questions.
- Memorise the Formula Recap first — every numerical depends on these six formulas.
- Work through sections in order: MCQs → Fill in the Blanks → True or False → Match the Following → Short Answers → Detail Answers.
- Check Common Mistakes to Avoid after every practice session to catch recurring errors.
- Use the Exam Checklist as a final review before the board exam.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Photoelectric effect | Emission of electrons from a metal surface when light of sufficient frequency falls on it |
| Work function (φ) | Minimum energy needed to release an electron from the metal surface |
| Threshold frequency (ν₀) | Minimum frequency of incident light required to cause photoelectric emission |
| Stopping potential (V₀) | Minimum reverse voltage needed to stop the fastest photoelectrons |
| Einstein's photoelectric equation | hν = φ + K_max; photon energy equals work function plus maximum kinetic energy |
| Photon | Quantum of electromagnetic radiation carrying energy E = hν |
| de Broglie wavelength | Wavelength associated with a moving particle: λ = h / p = h / (mv) |
| Matter waves | Waves associated with material particles, proposed by de Broglie |
| Photoelectric current | Current due to emitted photoelectrons; proportional to intensity of incident light |
| Saturation current | Maximum photoelectric current when all emitted electrons reach the collector |
| Emission spectrum | Set of frequencies of radiation emitted by a substance in excited state |
| Photoelectron | Electron emitted from a metal surface due to the photoelectric effect |
| Planck's constant (h) | Fundamental constant equal to 6.626 × 10⁻³⁴ J s; relates photon energy to frequency |
| Davisson-Germer experiment | Experiment that confirmed de Broglie's hypothesis by demonstrating electron diffraction |
Formula Recap
| Concept | Formula |
|---|
| Photon energy | E = h nu = hc / lambda |
| Einstein equation | h nu = phi + Kmax |
| Stopping potential | Kmax = eV0 |
| Threshold frequency | nu0 = phi / h |
| de Broglie wavelength | lambda = h / p |
| For electron | lambda = h / mv |
Photoelectric Numerical Pattern
1. Convert wavelength to metre or energy to joule/eV. 2. Find photon energy. 3. Use Einstein equation. 4. If stopping potential is asked, use Kmax = eV0. 5. Write result with correct unit.
I. Choose the Best Answer (MCQs)
1. The photoelectric effect proves which nature of light?
Answer
(a) Wave nature   (b) Particle nature   (c) Dual nature   (d) Electromagnetic nature
Answer(b) Particle nature — The photoelectric effect shows that light delivers energy in discrete quanta (photons), confirming the particle nature.
2. The work function of a metal is 4 eV. The threshold frequency is approximately:
Answer
(a) 9.66 × 10¹⁴ Hz   (b) 6.0 × 10¹⁴ Hz   (c) 1.21 × 10¹⁵ Hz   (d) 3.0 × 10¹⁴ Hz
Answer(a) 9.66 × 10¹⁴ Hz — ν₀ = φ/h = (4 × 1.6 × 10⁻¹⁹) / (6.626 × 10⁻³⁴) ≈ 9.66 × 10¹⁴ Hz.
3. If the intensity of incident light is doubled, the maximum kinetic energy of photoelectrons:
Answer
(a) Doubles   (b) Halves   (c) Remains the same   (d) Quadruples
Answer(c) Remains the same — K_max depends on frequency, not intensity. Doubling intensity increases the number of photoelectrons, not their maximum KE.
4. The de Broglie wavelength of a particle is inversely proportional to:
Answer
(a) Its mass only   (b) Its velocity only   (c) Its momentum   (d) Its energy
Answer(c) Its momentum — λ = h/p, so wavelength is inversely proportional to momentum (p = mv).
5. Which of the following has the longest de Broglie wavelength if all move with the same speed?
Answer
(a) Alpha particle   (b) Proton   (c) Electron   (d) Neutron
Answer(c) Electron — The lightest particle has the smallest momentum (p = mv) and hence the longest wavelength (λ = h/p).
6. In the Davisson-Germer experiment, the diffraction pattern confirmed that:
Answer
(a) Light has wave nature   (b) Electrons have particle nature   (c) Electrons have wave nature   (d) Photons have particle nature
Answer(c) Electrons have wave nature — The diffraction of electrons by a nickel crystal demonstrated their wave-like behaviour.
7. The stopping potential V₀ for a metal surface is 2 V. The maximum kinetic energy of photoelectrons is:
Answer
(a) 2 J   (b) 2 eV   (c) 3.2 × 10⁻¹⁹ J   (d) Both (b) and (c)
Answer(d) Both (b) and (c) — K_max = eV₀ = 2 eV = 2 × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J.
8. If the frequency of incident light is less than the threshold frequency, photoelectric emission:
Answer
(a) Increases slightly   (b) Does not occur   (c) Is independent of frequency   (d) Depends on intensity
Answer(b) Does not occur — Photoelectric emission is impossible if ν < ν₀, regardless of intensity.
9. The photon energy of light of wavelength 400 nm is approximately:
Answer
(a) 1.55 eV   (b) 3.1 eV   (c) 4.97 eV   (d) 6.2 eV
Answer(b) 3.1 eV — E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (400 × 10⁻⁹) ≈ 4.97 × 10⁻¹⁹ J ≈ 3.1 eV.
10. The de Broglie wavelength of an electron accelerated through a potential of V volts is:
Answer
(a) λ = h / √(2meV)   (b) λ = h / (2meV)   (c) λ = h√(2meV)   (d) λ = h² / (2meV)
Answer(a) λ = h / √(2meV) — Kinetic energy = eV = ½mv², so v = √(2eV/m) and λ = h/(mv) = h/√(2meV).
II. Fill in the Blanks
1. The minimum frequency required to cause photoelectric emission is called \_\_\_\_\_\_\_\_\_\_. (threshold frequency)
2. The maximum kinetic energy of photoelectrons is equal to \_\_\_\_\_\_\_\_\_\_. (eV₀)
3. According to de Broglie, the wavelength associated with a particle of momentum p is \_\_\_\_\_\_\_\_\_\_. (λ = h/p)
4. The work function of a metal is measured in \_\_\_\_\_\_\_\_\_\_. (electron-volt or joule)
5. In the photoelectric effect, the number of photoelectrons emitted per second is proportional to the \_\_\_\_\_\_\_\_\_\_ of incident light. (intensity)
6. The experiment that confirmed the wave nature of electrons was performed by \_\_\_\_\_\_\_\_\_\_. (Davisson and Germer)
7. If the stopping potential increases, the \_\_\_\_\_\_\_\_\_\_ kinetic energy of photoelectrons increases. (maximum)
8. The energy of a photon is proportional to its \_\_\_\_\_\_\_\_\_\_. (frequency)
III. True or False (with Corrections)
1. The photoelectric effect can occur at any frequency of incident light.
False. — Photoelectric emission occurs only when the frequency of incident light is greater than or equal to the threshold frequency (ν ≥ ν₀).
2. Increasing the intensity of light increases the maximum kinetic energy of photoelectrons.
False. — Increasing intensity increases the number of photoelectrons, not their maximum kinetic energy. K_max depends only on the frequency of incident light.
3. de Broglie wavelength is significant only for particles with very small mass.
True. — For macroscopic objects the de Broglie wavelength is negligibly small. It becomes measurable only for subatomic particles like electrons.
4. The stopping potential depends on the intensity of incident light.
False. — The stopping potential depends on the frequency of incident light and the work function of the metal, not on intensity.
5. A photon has energy but no rest mass.
True. — A photon carries energy E = hν and momentum p = h/λ, but its rest mass is zero.
6. In the Davisson-Germer experiment, the diffraction pattern was observed by varying the accelerating voltage.
True. — The intensity of scattered electrons was measured at different angles for different accelerating voltages, producing a diffraction pattern.
7. The saturation current depends on the frequency of incident light.
False. — The saturation current depends on the intensity of incident light. Higher intensity produces more photoelectrons, giving a larger saturation current.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Work function | (a) λ = h / √(2meV) |
| 2. Stopping potential | (b) Minimum energy to release electron |
| 3. de Broglie wavelength (electron) | (c) Maximum kinetic energy of photoelectrons |
| 4. Einstein's photoelectric equation | (d) hν = φ + K_max |
| 5. Threshold frequency | (e) ν₀ = φ / h |
| 6. Photon energy | (f) E = hν |
Answer1 → (b), 2 → (c), 3 → (a), 4 → (d), 5 → (e), 6 → (f)
V. Answer Briefly (2-mark)
1. Define work function. Does it depend on the incident light?
Answer
The work function (φ) is the minimum energy required to liberate an electron from the surface of a metal. It is a property of the material and depends on the nature of the metal surface. It does not depend on the incident light; it remains constant for a given metal regardless of the frequency or intensity of light.
2. What is the significance of the threshold frequency in the photoelectric effect?
Answer
The threshold frequency (ν₀) is the minimum frequency of incident radiation below which no photoelectric emission occurs, no matter how high the intensity. It relates to the work function by ν₀ = φ/h. Light with frequency below ν₀ cannot supply enough energy per photon to overcome the work function, so emission is impossible.
3. Why do macroscopic objects not show noticeable wave properties?
Answer
The de Broglie wavelength λ = h/(mv) is extremely small for macroscopic objects because their mass is very large. Since h (6.626 × 10⁻³⁴ J s) is tiny, dividing by the large momentum of a macroscopic body gives a wavelength far too small to detect. Only subatomic particles like electrons have measurable de Broglie wavelengths.
4. State the relation between stopping potential and maximum kinetic energy.
Answer
The stopping potential (V₀) is the minimum reverse voltage that stops the fastest photoelectrons. The maximum kinetic energy equals the work done by this voltage: K_max = eV₀, where e is the electron charge. This relation allows us to determine K_max experimentally by measuring V₀.
5. What observation in the Davisson-Germer experiment confirmed de Broglie's hypothesis?
Answer
Davisson and Germer observed a peak in the intensity of scattered electrons at a specific angle (50°) when the accelerating voltage was 54 V. This peak matched the diffraction condition for a wavelength equal to the de Broglie wavelength of the electrons, confirming that electrons exhibit wave-like behaviour.
VI. Answer in Detail (5-mark)
1. Explain Einstein's photoelectric equation and its significance.
Answer
Einstein extended Planck's quantum theory to explain the photoelectric effect. He proposed that light consists of discrete packets of energy called photons, each carrying energy E = hν. When a photon strikes a metal surface, its energy is transferred entirely to a single electron. If the photon energy (hν) exceeds the work function (φ) of the metal, the electron is emitted. The surplus energy becomes the maximum kinetic energy of the photoelectron. This gives the equation: hν = φ + K_max. The significance of this equation is immense. First, it explains why a threshold frequency exists: if hν < φ, no emission occurs regardless of intensity. Second, it explains why K_max depends on frequency and not intensity — increasing intensity sends more photons but each photon still carries the same energy hν. Third, it explains the instantaneous nature of the emission — a single photon ejects a single electron in one event, requiring no energy accumulation time. Einstein's equation earned him the Nobel Prize in 1921 and laid the foundation for quantum mechanics.
2. Describe the experimental observations of the photoelectric effect and how they contradict wave theory.
Answer
The photoelectric effect was first observed by Hertz and later studied systematically by Hallwachs, Lenard, and others. The key experimental observations are: (i) For a given metal, photoelectric emission occurs only if the frequency of incident light exceeds a threshold value ν₀, irrespective of intensity. (ii) The maximum kinetic energy of photoelectrons increases linearly with frequency but is independent of intensity. (iii) The photoelectric current (number of photoelectrons per second) is directly proportional to the intensity of incident light. (iv) Photoelectric emission is instantaneous — there is no time lag between illumination and emission, even at very low intensities. (v) Increasing the intensity of light below the threshold frequency never produces emission. Wave theory contradicts all these observations. According to wave theory, energy is distributed uniformly over the wavefront, so increasing intensity should increase K_max — but experimentally K_max depends on frequency, not intensity. Wave theory also predicts a time lag for electrons to accumulate energy at low intensities, yet emission is instantaneous. Wave theory cannot explain the threshold frequency either. Einstein's photon model resolved every contradiction by treating light as quanta.
3. State and explain the de Broglie hypothesis. Describe the Davisson-Germer experiment.
Answer
De Broglie hypothesised that if light exhibits both wave and particle nature, then material particles should also exhibit wave nature. He proposed that a particle of momentum p has an associated wavelength λ = h/p, where h is Planck's constant. For an electron of mass m moving with velocity v, the de Broglie wavelength is λ = h/(mv). If the electron is accelerated through a potential V, then eV = ½mv², and the wavelength becomes λ = h/√(2meV). The Davisson-Germer experiment (1927) confirmed this hypothesis. In their setup, electrons were accelerated through a known potential and directed at a nickel crystal. The scattered electrons were collected by a detector that could move in an arc to measure intensity at different angles. When the accelerating voltage was 54 V, a pronounced peak in intensity was observed at a scattering angle of 50°. Using Bragg's law for diffraction, the experimental wavelength matched the de Broglie wavelength of electrons at that voltage (approximately 1.67 Å). This demonstrated that electrons undergo diffraction just like waves, confirming de Broglie's matter-wave hypothesis.
4. Explain stopping potential and threshold frequency. How are they related?
Answer
Stopping potential (V₀) is the minimum reverse voltage applied between the anode and cathode that just stops the most energetic photoelectrons from reaching the collector. At this voltage the photoelectric current drops to zero. The maximum kinetic energy of photoelectrons equals the work done by the stopping potential: K_max = eV₀. Threshold frequency (ν₀) is the minimum frequency of incident radiation below which no photoelectric emission occurs, regardless of the light's intensity. It is related to the work function by ν₀ = φ/h. These two quantities are related through Einstein's equation. When ν = ν₀, the photon energy exactly equals the work function, and K_max = 0, which means V₀ = 0. For ν > ν₀, we get K_max = h(ν − ν₀) and therefore eV₀ = h(ν − ν₀), or V₀ = (h/e)(ν − ν₀). This linear relation between V₀ and ν is a key result confirmed experimentally. The slope of the V₀ versus ν graph gives h/e, and the intercept on the frequency axis gives ν₀. From these, both Planck's constant and the work function can be determined experimentally.
Short Answer Points
- Photoelectric effect supports particle nature of light.
- Threshold frequency depends on material.
- Increasing intensity increases number of emitted electrons, not maximum kinetic energy.
- de Broglie hypothesis says matter has wave nature.
Worked Examples
Example 1 — Photon energy
Find the energy of a photon of wavelength 500 nm (in joules and eV).
E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸) / (500×10⁻⁹)
= 1.989×10⁻²⁵ / 5×10⁻⁷ = 3.978×10⁻¹⁹ J
In eV: E = 3.978×10⁻¹⁹ / 1.6×10⁻¹⁹ ≈ 2.49 eV
Example 2 — Photoelectric effect
Light of energy 5 eV strikes a metal of work function 2 eV. Find the maximum KE of emitted electrons.
KE_max = hf − φ = 5 − 2 = 3 eV
Example 3 — de Broglie wavelength
Find the de Broglie wavelength of an electron (m = 9.1×10⁻³¹ kg) moving at 10⁶ m/s.
λ = h / (mv) = (6.63×10⁻³⁴) / (9.1×10⁻³¹ × 10⁶)
= 6.63×10⁻³⁴ / 9.1×10⁻²⁵ ≈ 7.3×10⁻¹⁰ m
Common Mistakes to Avoid
1. ❌ Confusing frequency and intensity effects ✅ Intensity affects the number of photoelectrons (photoelectric current). Frequency affects their maximum kinetic energy. Never swap these.
2. ❌ Not converting eV to joule (or vice versa) ✅ Always check the unit asked in the question. Use 1 eV = 1.6 × 10⁻¹⁹ J. Write the conversion step explicitly to avoid calculation errors.
3. ❌ Using wrong value of h or c ✅ Memorise: h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s. For energy in eV, use h = 4.136 × 10⁻¹⁵ eV s.
4. ❌ Forgetting the threshold condition ✅ Always check ν ≥ ν₀ before solving. If ν < ν₀, photoelectric emission does not occur — no calculation is needed.
5. ❌ Assuming de Broglie wavelength is significant for all objects ✅ λ = h/p is negligible for macroscopic bodies. Apply it only to subatomic particles (electrons, protons, neutrons).
6. ❌ Writing K_max = eV₀ without verifying which plate is positive ✅ Stopping potential is applied in reverse — collector is negative relative to emitter. K_max = eV₀ is always valid, but remember V₀ is the magnitude of the stopping voltage.
Exam Checklist
- [ ] Work function (φ) is handled correctly — know its value in eV or J as needed.
- [ ] Unit conversion between eV and joule is shown explicitly.
- [ ] Threshold frequency ν₀ = φ/h is calculated before checking emission condition.
- [ ] Einstein's equation hν = φ + K_max is applied step by step.
- [ ] Stopping potential relation K_max = eV₀ is used correctly.
- [ ] Intensity vs frequency distinction is stated in conceptual answers.
- [ ] de Broglie wavelength formula λ = h/p is applied with correct momentum p = mv.
- [ ] For accelerated electrons, use λ = h/√(2meV) — memorise or derive it.
- [ ] Davisson-Germer experiment details (54 V, 50°, nickel crystal) are recalled for long answers.
- [ ] Final answer always carries the correct unit (J, eV, m, Hz) and appropriate significant figures.
