ટોપિક 8 / 27

Electron Emission Photoelectric

🎓 Class 12 Physics CBSE Theory Ch 11 – Dual Nature of Radiation and Matter ⏱ ~14 min
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Electron Emission Photoelectric

આ મૂલ્યાંકન આના પર આધારિત હશે: Electron Emission Photoelectric

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Electron Emission Photoelectric

11.1 Introduction — Two Faces of Light and Matter

By the close of the nineteenth century, classical physics seemed complete: Newton's mechanics ruled the world of particles and Maxwell's equations ruled the world of electromagnetic waves. Then a series of experiments — discharge tubes, X-rays, the photoelectric effect — exposed cracks in this neat division. Light, long considered a wave, sometimes behaves like a stream of particles. And electrons, long considered particles, sometimes behave like waves. This chapter tells the story of how these two faces were uncovered, beginning with the simplest question of all — how do electrons leave a metal?

Free electrons: A metal contains a "sea" of nearly free electrons that drift through the lattice when an electric field is applied (this is what gives a metal its conductivity). At ordinary temperatures, however, these electrons cannot escape the surface. Why? Because the positive ion-cores attract them back, creating an invisible potential barrier.

11.2 Electron Emission & the Work Function

Imagine a free electron sitting just inside a metal. To break free into the surrounding vacuum, it must perform a small but definite amount of work against the attractive pull of the positive ions left behind. The minimum energy needed for this escape is the work function of the metal, denoted \(\phi_0\).

Work function \(\phi_0\): the minimum energy required to liberate a free electron from a metal surface. It is measured in joules but is more conveniently expressed in electron-volts (eV), where \(1\,\text{eV} = 1.602\times10^{-19}\,\text{J}\).

The work function depends on the metal and on the cleanliness of its surface. A few representative values:

Metal\(\phi_0\) (eV)Metal\(\phi_0\) (eV)
Caesium (Cs)2.14Aluminium (Al)4.28
Potassium (K)2.30Mercury (Hg)4.49
Sodium (Na)2.75Copper (Cu)4.65
Calcium (Ca)3.20Silver (Ag)4.70
Molybdenum (Mo)4.17Platinum (Pt)5.65

Table 11.1 — Work functions of common metals (NCERT Table 11.1).

Four Ways to Free an Electron

The energy needed to overcome \(\phi_0\) can be supplied in four distinct ways, giving four types of electron emission:

  1. Thermionic emission — Heating the metal raises the kinetic energy of free electrons until some can escape. Used in cathode-ray tubes and old vacuum-tube valves.
  2. Field (cold) emission — A very strong electric field (~108 V·m−1), applied to a sharp metal tip, pulls electrons out without heating. Used in spark plugs and field-emission microscopes.
  3. Photoelectric emission — Light of suitable frequency striking the surface ejects electrons. The freed particles are called photoelectrons.
  4. Secondary emission — Fast-moving electrons or ions, on hitting the metal, knock out further electrons (used in photomultipliers).
Thermionic Heat (T high) Field Strong E-field Photoelectric Light (hν) Secondary Fast e or ion impact e
Fig 11.1: Four mechanisms of electron emission. Red dots are the ejected electrons.

11.3 Photoelectric Effect — Hertz's Discovery

In 1887, while studying spark discharges between metal balls connected to a high-voltage source, Heinrich Hertz made an accidental but profound observation: when ultraviolet light from one spark fell on the negative terminal of his second spark gap, the second spark fired more easily. Light, somehow, was helping electrons leave the metal. This was the photoelectric effect.

Hallwachs and Lenard (1886–1902)

Wilhelm Hallwachs and Philipp Lenard investigated the effect more carefully:

  • A negatively charged zinc plate illuminated with UV light lost its charge — clearly negative carriers were being emitted.
  • A neutral zinc plate became positively charged under UV light, then accumulated more positive charge as more electrons left.
  • A positively charged plate accumulated even more positive charge under UV — but only because emitted electrons were quickly attracted back to the surface.
  • The effect occurred only above a certain frequency, called the threshold frequency, characteristic of the metal.

Lenard further showed that the emitted particles had the same charge-to-mass ratio as cathode rays — they were the same electrons J. J. Thomson had discovered in 1897. Among alkali metals (sodium, potassium, caesium, rubidium), even visible light was enough to release electrons; with most other metals, ultraviolet light was required.

UV lamp UV photons Zn plate (initially neutral) Electroscope leaves diverge e
Fig 11.2: Hallwachs/Lenard setup. UV light ejects electrons from the zinc plate; the connected gold-leaf electroscope reveals the loss of negative charge.

11.4 Experimental Study of the Photoelectric Effect

To make the photoelectric effect quantitative, an evacuated glass tube houses a clean metal photo-emitter (the cathode C) and a metal collector (the anode A). Monochromatic light of frequency \(\nu\) falls on C through a quartz window. A battery sets up a potential difference between A and C; a sensitive microammeter measures the current. The setup allows three independent variables to be varied:

  1. Intensity \(I\) of the incident light (at fixed \(\nu\) and fixed plate voltage).
  2. Frequency \(\nu\) (at fixed \(I\) and fixed potential).
  3. Potential difference \(V\) between A and C, including reversal of polarity.
Light C cathode photoelectrons A anode Battery μA V
Fig 11.3: Schematic of the apparatus used to study the photoelectric effect — evacuated tube, cathode C, anode A, variable potential, microammeter and voltmeter.

11.4.1 Effect of Intensity (constant \(\nu\), constant \(V\))

When the frequency exceeds the threshold, the photoelectric current \(I\) is found to be directly proportional to the intensity of the incident light. Doubling the brightness doubles the number of electrons released per second. A straight-line graph through the origin summarises the result.

11.4.2 Effect of Potential (constant \(\nu\), constant \(I\))

For a positive accelerating potential on A, the current rises with \(V\) and then saturates — every photoelectron emitted now reaches the anode. If the potential on A is reversed (made negative — a "retarding" voltage), the current decreases. At a sharp, particular value \(V_0\), called the stopping potential, the current falls to zero. The work done by the field exactly equals the largest kinetic energy of the photoelectrons:

eV0 = (½)m v2max = Kmax

Increasing the intensity shifts the saturation level upward, but the stopping potential is the same. The maximum kinetic energy of photoelectrons therefore depends on light quality (frequency), not quantity (intensity).

V (anode potential) Photocurrent I −V₀ 0 I₃ (high) I₂ (mid) I₁ (low) Same V₀ for all intensities
Fig 11.4: Photoelectric current vs anode potential for three intensities at the same frequency. Saturation depends on intensity but the stopping potential V₀ is identical.

11.4.3 Effect of Frequency (constant \(I\))

When the experiment is repeated at three different frequencies of incident light (each above the threshold), three new \(I\)-vs-\(V\) curves emerge. Each saturates at the same value (intensity unchanged), but the stopping potential becomes more negative as frequency increases. Plotting \(V_0\) against \(\nu\) gives a perfect straight line, intersecting the \(\nu\)-axis at the threshold frequency \(\nu_0\). Below \(\nu_0\), no photoelectrons are released, no matter how intense the light is.

ν V₀ ν₀ (threshold) slope = h/e V₀ = (h/e)ν − φ₀/e
Fig 11.5: Stopping potential V₀ versus frequency ν. The intercept on the ν-axis gives the threshold frequency ν₀; the slope equals h/e (Millikan's measurement).
Three experimental laws of the photoelectric effect:
  1. For a given metal and frequency, the saturation current is proportional to intensity.
  2. For a given metal, there is a threshold frequency below which no emission occurs, however intense the light.
  3. Above threshold, the maximum kinetic energy of photoelectrons increases linearly with frequency, independent of intensity.
And — crucially — the photoelectric process is essentially instantaneous (lag < 10−9 s), even at very low light intensity.
Activity 11.1 — Charging by Light

You will need: a freshly cleaned zinc plate, a gold-leaf (or simple paper-strip) electroscope, an ebonite rod with woollen cloth, and a UV source (a desk-lamp UV-A bulb works for a clean Zn surface).

Predict: If you charge the zinc plate negatively and then shine UV on it, what will the leaves of the electroscope do?
  1. Connect the zinc plate to the cap of the electroscope.
  2. Charge the assembly negatively by stroking with the ebonite rod.
  3. The leaves diverge because they share the same negative charge.
  4. Now switch on the UV light. Observe the leaves collapse.
  5. Repeat with the lamp shielded by an ordinary glass plate (which absorbs UV). The collapse stops.
UV photons have enough energy (\(h\nu > \phi_0\) for zinc \(\phi_0\approx4.3\) eV) to eject electrons from the zinc surface. As negative charge is removed, the leaves come together. Visible light (after the glass barrier) lacks the required frequency, so the leaves stay apart — direct evidence of a frequency threshold.

Interactive — Photoelectric Apparatus

Slide the intensity to set how many photons strike the cathode each second; slide the anode potential through positive and negative values. Watch the current curve update and notice that the stopping potential (red marker) does not shift with intensity.

5
+1.0 V
V I 0 −V₀

Frequency is held above threshold so V₀ is fixed. Intensity rescales the saturation height; the operating point moves along the chosen curve.

Worked Examples

Example 1 — Threshold wavelength of caesium

The work function of caesium is \(\phi_0 = 2.14\) eV. Find the threshold frequency and threshold wavelength for photoelectric emission.

\(\nu_0 = \dfrac{\phi_0}{h} = \dfrac{2.14\times1.602\times10^{-19}}{6.626\times10^{-34}} = 5.17\times10^{14}\) Hz.
\(\lambda_0 = \dfrac{c}{\nu_0} = \dfrac{3\times10^8}{5.17\times10^{14}} \approx 5.80\times10^{-7}\) m \(=580\) nm — yellow visible light.
Example 2 — Why copper needs UV but caesium does not

Compare the threshold wavelengths of copper (\(\phi_0=4.65\) eV) and caesium (\(\phi_0=2.14\) eV). Which one responds to ordinary visible light?

For Cu: \(\lambda_0 = hc/\phi_0 = 1242/4.65 \approx 267\) nm — far ultraviolet.
For Cs: \(\lambda_0 = 1242/2.14 \approx 580\) nm — yellow.
Caesium photo-emits with any visible light bluer than yellow; copper needs UV.
Example 3 — Photoelectrons per second

Light of intensity \(2\times10^{-6}\) W·m−2 and wavelength 500 nm strikes a 1 cm² photocathode. If 1 in every 1000 photons ejects an electron, find the photocurrent.

Power on cathode \(P = 2\times10^{-6}\times10^{-4} = 2\times10^{-10}\) W.
Energy per photon \(E = hc/\lambda = 1.99\times10^{-25}/500\times10^{-9} = 3.98\times10^{-19}\) J.
Photons/s \(= P/E = 5.03\times10^{8}\). Electrons/s = 5.03\times10^{5}\). Current \(I = ne = 5.03\times10^{5}\times1.6\times10^{-19} = 8.05\times10^{-14}\) A \(\approx 81\) fA.

Competency-Based Questions

Q1. The minimum energy required by an electron to escape from a metal surface is called the:

  • (a) Ionisation energy
  • (b) Work function
  • (c) Binding energy
  • (d) Stopping energy
(b) The work function \(\phi_0\) is the threshold energy needed to liberate an electron from the metal surface.

Q2. Which of the following emissions occurs when a sharp metal tip is placed in a very strong electric field?

  • (a) Thermionic
  • (b) Photoelectric
  • (c) Field emission
  • (d) Secondary emission
(c) Field emission — fields ~108 V/m extract electrons by tunnelling through the surface barrier.

Q3. (Short Answer) Why does a positively charged plate gain even more positive charge when illuminated with UV?

UV photons eject photoelectrons. Most are pulled back by the positive plate, but the small fraction with high enough kinetic energy escape, reducing the negative count and so increasing the net positive charge.

Q4. (True/False) Increasing the intensity of light always increases the maximum kinetic energy of the emitted photoelectrons.

False. Intensity affects the number of photoelectrons (and hence saturation current); the maximum kinetic energy depends only on frequency.

Q5. (HOT) When monochromatic light of frequency \(\nu>\nu_0\) is shone on a clean metal, the stopping potential is \(V_0\). If the intensity is doubled but the frequency unchanged, what happens to (i) saturation current, (ii) stopping potential?

(i) Saturation current doubles (more photons → more photoelectrons). (ii) Stopping potential remains \(V_0\) — the maximum KE per electron is set by \(h\nu - \phi_0\), which has not changed.

Assertion–Reason Questions

Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion: Alkali metals like caesium are commonly used in photocells.

Reason: Their work functions are low enough for visible light to liberate electrons.

(A) Both statements are correct, and the reason explains the assertion. Cs has \(\phi_0=2.14\) eV, allowing yellow-green light to trigger the photoelectric effect.

Assertion: Photoelectric emission shows no time-lag, even at very low light intensity.

Reason: Each photoelectron absorbs the entire energy of a single photon in one go.

(A) Both true and the reason is the proper explanation — the one-photon-one-electron interaction is essentially instantaneous.

Assertion: Below the threshold frequency, no electrons are emitted from a metal even if a very intense beam is used.

Reason: A high-intensity beam carries more photons but each photon still has the same insufficient energy.

(A) Both correct. The energy budget is per photon, not summed over photons.

Frequently Asked Questions - Electron Emission Photoelectric

What is the main concept covered in Electron Emission Photoelectric?
In NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter), "Electron Emission Photoelectric" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Electron Emission Photoelectric useful in real-life applications?
Real-life applications of "Electron Emission Photoelectric" from NCERT Class 12 Physics Chapter 11 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Electron Emission Photoelectric?
Key formulas in "Electron Emission Photoelectric" (NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 11?
NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter) is structured so each part builds on the previous one. "Electron Emission Photoelectric" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Electron Emission Photoelectric?
CBSE board questions from "Electron Emission Photoelectric" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Electron Emission Photoelectric" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
AI ટ્યુટર
Physics Class 12 Part II – NCERT (2025-26)
તૈયાર
નમસ્તે! 👋 હું ગૌરા છું, Electron Emission Photoelectric માટે તમારું AI ટ્યુટર. આરામથી પાઠ ભણો — જ્યારે પણ કોઈ શંકા થાય, બસ મને પૂછો! હું મદદ માટે અહીં જ છું.

🎯 Physics ની પ્રેક્ટિસ કરો

તમે જે ભણ્યા તેનું પૂરું પેપર આપો, પ્રશ્ન દીઠ તપાસાયેલું.

બોર્ડ પરીક્ષા સેમ્પલ પેપર

આ વિષયનાં બધાં પેપર →

મોક પરીક્ષાઓ

આ વિષયની બધી મોક પરીક્ષાઓ →

🎁 Join our community and get free AI credits!