This MCQ module is based on: Electron Emission Photoelectric
Electron Emission Photoelectric
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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?
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 \(\phi_0\).
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) | 2.14 | Aluminium (Al) | 4.28 |
| Potassium (K) | 2.30 | Mercury (Hg) | 4.49 |
| Sodium (Na) | 2.75 | Copper (Cu) | 4.65 |
| Calcium (Ca) | 3.20 | Silver (Ag) | 4.70 |
| Molybdenum (Mo) | 4.17 | Platinum (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:
- 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.
- 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.
- Photoelectric emission — Light of suitable frequency striking the surface ejects electrons. The freed particles are called photoelectrons.
- Secondary emission — Fast-moving electrons or ions, on hitting the metal, knock out further electrons (used in photomultipliers).
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.
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:
- Intensity \(I\) of the incident light (at fixed \(\nu\) and fixed plate voltage).
- Frequency \(\nu\) (at fixed \(I\) and fixed potential).
- Potential difference \(V\) between A and C, including reversal of polarity.
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:
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).
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.
- For a given metal and frequency, the saturation current is proportional to intensity.
- For a given metal, there is a threshold frequency below which no emission occurs, however intense the light.
- Above threshold, the maximum kinetic energy of photoelectrons increases linearly with frequency, independent of intensity.
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).
- Connect the zinc plate to the cap of the electroscope.
- Charge the assembly negatively by stroking with the ebonite rod.
- The leaves diverge because they share the same negative charge.
- Now switch on the UV light. Observe the leaves collapse.
- Repeat with the lamp shielded by an ordinary glass plate (which absorbs UV). The collapse stops.
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.
Frequency is held above threshold so V₀ is fixed. Intensity rescales the saturation height; the operating point moves along the chosen curve.
Worked Examples
The work function of caesium is \(\phi_0 = 2.14\) eV. Find the threshold frequency and threshold wavelength for photoelectric emission.
\(\lambda_0 = \dfrac{c}{\nu_0} = \dfrac{3\times10^8}{5.17\times10^{14}} \approx 5.80\times10^{-7}\) m \(=580\) nm — yellow visible light.
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 Cs: \(\lambda_0 = 1242/2.14 \approx 580\) nm — yellow.
Caesium photo-emits with any visible light bluer than yellow; copper needs UV.
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.
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:
Q2. Which of the following emissions occurs when a sharp metal tip is placed in a very strong electric field?
Q3. (Short Answer) Why does a positively charged plate gain even more positive charge when illuminated with UV?
Q4. (True/False) Increasing the intensity of light always increases the maximum kinetic energy of the emitted photoelectrons.
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?
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.
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.
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.
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Board exam sample papers
Physics — CBSE Class XII Sample Paper 1 (2025-26)
Section A · Section B · Section C · Section D · Section E