Electron emission

Hold a piece of copper wire in your hand and, without you feeling a thing, billions of free electrons inside it are darting about at enormous speed — yet not one of them drifts out into the air. Something is holding them in, invisibly, the way the rim of a bowl holds a marble rolling around inside it. That invisible wall at the metal's surface is the starting puzzle of the whole chapter "Dual Nature of Radiation and Matter": what keeps electrons trapped inside a metal, and what does it take to set them free? The answer turns out to need only one number per metal — a minimum energy called the work function — and three different ways of paying that energy bill: heat the metal, apply an intense electric field, or shine light of the right kind on it. That third route, photoelectric emission, is the one this chapter builds its whole argument around, because in 1905 Einstein used it to show that light itself sometimes behaves like a stream of particles, not just a wave. Before that story can be told, though, you need to be completely at home with the idea of a surface energy barrier, its size in electron volts, and how it converts into a threshold frequency of light — the exact vocabulary this article sets up. We will look at why the barrier exists physically, meet the three emission mechanisms by name, derive the eV-to-joule and work-function-to-threshold-frequency relations you will use in every problem for the rest of the chapter, and work through two numerical examples — one board-style, one competitive-exam style — using real work-function values for common metals. By the end you should be able to look at a work function in electron volts and immediately state the minimum frequency (or maximum wavelength) of light that could liberate an electron from that metal, no calculator surprises included.

Key takeaways

Electrons are trapped, not free to leave. Conduction electrons roam inside a metal but are held back at the surface by the net attractive pull of the positive ion lattice — an energy barrier, not a physical wall.
The work function $\phi_0$ is the barrier's height. It is the minimum energy needed to just free the least tightly bound electron from a metal surface, usually quoted in electron volts (eV), typically $2\ \text{eV}$ to $6\ \text{eV}$ for common metals.
Three mechanisms pay this energy bill. Thermionic emission (heat), field emission (a strong external electric field), and photoelectric emission (light) — each supplies $\phi_0$ by a different physical route.
Photoelectric emission is quantum, not classical. A single photon of energy $h\nu$ either has enough energy to free one electron or it does not — this one-photon-one-electron picture is what the rest of the chapter builds on.
Work function fixes a threshold frequency. Setting $h\nu_0 = \phi_0$ gives the minimum frequency $\nu_0$ of light that can, in principle, cause emission from a given metal — below $\nu_0$ , no amount of light intensity helps.
Unit conversion is not optional. $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ must be applied correctly before combining $\phi_0$ with $h$ in SI units — this single slip accounts for most marks lost on this topic.
Key idea: a metal traps its free electrons behind a surface energy barrier of height equal to the work function $\phi_0$ , and thermionic, field, or photoelectric emission are simply three different ways of supplying that energy, with $\phi_0 = h\nu_0$ fixing the threshold frequency for the photoelectric route.

The core insight

Free electrons inside a metal are held back by a surface potential-energy barrier of height $\phi_0$ (the work function), and thermionic, field, and photoelectric emission are three distinct physical mechanisms — heat, external electric field, and absorbed light quanta respectively — for supplying at least $\phi_0$ to let an electron escape, with $\phi_0 = h\nu_0$ defining the threshold frequency for light.

Where it fits in the syllabus

Electron emission opens Unit IX, "Dual Nature of Radiation and Matter," in the CBSE Class 12 Physics syllabus, and everything else in the chapter leans on the vocabulary set up here. The prescribed scope keeps this topic firmly qualitative and historical — Hertz's and Lenard's observations, the three emission methods, and the work function as a fixed metal property — with the real quantitative payoff arriving next, in Einstein's photoelectric equation $eV_0 = h\nu - \phi_0$ . Nothing here goes beyond what the board expects: no derivation of the barrier from first principles, no band-theory treatment of why the barrier exists at the level of solid-state physics — just the energetic picture and its direct numerical consequences.

The physical picture

Picture the free electrons of a metal as a crowd of very fast, very light particles confined to a room whose walls are not solid but are made of an invisible, uphill energy slope. Deep inside the metal an electron can wander freely at high kinetic energy, but as it approaches the surface, the net attraction of the positively charged ion lattice behind it (no longer cancelled by ions on the outside, since there are none there) pulls it back — exactly like a ball losing speed as it is thrown up a hill. Only an electron that arrives at the surface with enough kinetic energy to "climb" over this hill actually escapes into the vacuum outside.

\phi_0 = \text{minimum energy needed to just free the least strongly bound electron from a metal surface, at 0 K}

where: $\phi_0$ , the work function, is measured in joules or (far more conveniently for atomic-scale numbers) electron volts. One electron volt is defined as the kinetic energy gained by one electron accelerated through a potential difference of exactly one volt, and the conversion is fixed and exact enough for all school-level use:

1\ \text{eV} = 1.6\times10^{-19}\ \text{J}

The physical reading of $\phi_0$ is simple: it is a property of the metal and its surface condition — not of any light or field applied to it. Two samples of the same metal can even have slightly different work functions if one surface is oxidised or contaminated, because the surface condition changes the shape of the energy barrier. Typical values run from about $2.14\ \text{eV}$ for caesium (a very "loosely held" metal, useful in photocells) up to around $5.65\ \text{eV}$ for platinum (a much "tighter" metal). Once you accept that $\phi_0$ is a fixed energy cost, the three emission mechanisms are just three different ways of paying it:

E_{\text{supplied}} \geq \phi_0 \implies \text{electron escapes}

where: $E_{\text{supplied}}$ is the energy delivered to a particular electron by whichever mechanism is at work — thermal agitation, an external electric field doing work on the electron, or a single absorbed photon.

Figure

A metal block with electrons trapped below a hill shaped energy barrier at the surface, with three labelled arrows showing heat, a strong electric field, and a light photon each pushing an electron up and over the barrier

The three named mechanisms, in the order the syllabus lists them:

Thermionic emission heats the metal to a high temperature (as in the hot filament of an old vacuum-tube valve), so that a fraction of the free electrons pick up enough thermal kinetic energy, from random collisions, to exceed $\phi_0$ and escape. Only the high-energy tail of the thermal distribution manages this, which is why appreciable thermionic emission needs temperatures of a thousand kelvin or more.

Field emission applies an extremely strong external electric field (of order $10^8\ \text{V/m}$ ) right at the metal's surface. This field does work on an electron as it tries to leave and effectively narrows and lowers the energy barrier enough that electrons can escape even without extra thermal or optical energy — a process closely related to quantum tunnelling, though the syllabus keeps it at a qualitative level.

Photoelectric emission is the one this chapter is really about: light of suitable frequency falls on the metal surface, and the energy arrives not as a smooth continuous wave but in discrete parcels — photons — each carrying energy $E = h\nu$ , where $h$ is Planck's constant ( $6.63\times10^{-34}\ \text{J s}$ ) and $\nu$ is the light's frequency. A single photon can be absorbed by a single electron; if that photon's energy exceeds $\phi_0$ , the electron escapes, with any leftover energy becoming its kinetic energy outside the metal. This one-photon-one-electron rule — not the total brightness of the beam — is what governs whether emission happens at all, and it is the seed from which Einstein's photoelectric equation grows in the next article.

Derivation AFrom the work function to the threshold frequency

The goal is to turn the single measured quantity $\phi_0$ (the work function, usually given in eV) into $\nu_0$ , the threshold frequency — the minimum frequency of light that can, in principle, cause photoelectric emission from a given metal — using nothing beyond the photon energy relation $E = h\nu$ and careful unit conversion between electron volts and joules; this is the exact calculation every board and competitive question on this topic ultimately reduces to.

Step 1. Start from the photon energy relation. A photon of frequency $\nu$ carries energy $E = h\nu$ , with $h = 6.63\times10^{-34}\ \text{J s}$ and $\nu$ in hertz, giving $E$ in joules. This is the quantum (particle-like) description of light that the chapter is built around, in contrast to the classical wave picture where a beam's energy depends only on its intensity.

Step 2. Define the threshold condition. The threshold frequency $\nu_0$ is, by definition, the frequency at which a single photon carries exactly enough energy to overcome the work function and free an electron with zero leftover kinetic energy: $h\nu_0 = \phi_0$ . Any photon with $\nu < \nu_0$ carries less energy than $\phi_0$ and cannot cause emission, no matter how many such photons arrive per second.

Step 3. Solve for $\nu_0$ in terms of $\phi_0$ . Rearranging gives $\nu_0 = \dfrac{\phi_0}{h}$ . Because $h$ is a universal constant, this equation says the threshold frequency depends only on the work function — a property of the metal — and on nothing about the light source itself.

Step 4. Handle the units correctly. Work functions are conventionally quoted in electron volts, but $h$ is defined in joule-seconds, so $\phi_0$ must first be converted using $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ : if $\phi_0$ is given as $\phi_0^{\text{(eV)}}$ electron volts, then $\phi_0 = \phi_0^{\text{(eV)}}\times1.6\times10^{-19}\ \text{J}$ , and only this joule value should be divided by $h$ . Skipping this conversion, or dividing the eV number directly by $h$ , is the single most common source of a wrong final answer on this topic.

Step 5. Read the physical meaning. Because $\nu_0 = \phi_0/h$ is a straight proportionality, a metal with a larger work function always has a larger threshold frequency — it needs "bluer," higher-energy light to begin emitting at all. This proportionality is exactly what is used, and extended with the full stopping-potential relation, in Einstein's photoelectric equation covered in the next article.

Worked examples

Board level - threshold frequency of sodium

Problem: The work function of sodium is $2.3\ \text{eV}$ . Express this in joules, and find the threshold frequency of light for sodium.

Step 1. Convert the work function to joules using $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ :

\phi_0 = 2.3\times1.6\times10^{-19}\ \text{J} = 3.68\times10^{-19}\ \text{J}

Step 2. Apply the threshold condition $h\nu_0 = \phi_0$ and solve for $\nu_0$ , using $h = 6.63\times10^{-34}\ \text{J s}$ :

\nu_0 = \frac{\phi_0}{h} = \frac{3.68\times10^{-19}}{6.63\times10^{-34}} = 5.55\times10^{14}\ \text{Hz}

Step 3. State the result with its meaning. Sodium's threshold frequency is $\boxed{\nu_0 \approx 5.55\times10^{14}\ \text{Hz}}$ — this lies in the visible spectrum (roughly the blue-green region), which is exactly why sodium is a metal in which visible light can trigger photoelectric emission, unlike metals with a much higher work function.

Competitive level - comparing two metals and a mixed-light source

Problem: Metal A has work function $4.5\ \text{eV}$ and metal B has work function $2.1\ \text{eV}$ . A light source emits two frequencies simultaneously, $\nu_1 = 6.0\times10^{14}\ \text{Hz}$ and $\nu_2 = 1.2\times10^{15}\ \text{Hz}$ . For each metal, state which of these frequencies (if any) can cause photoelectric emission, and find the ratio of the two metals' threshold frequencies.

Step 1. Find the threshold frequency of each metal from $\nu_0 = \phi_0/h$ , converting eV to joules first. For metal A: $\phi_{0A} = 4.5\times1.6\times10^{-19} = 7.20\times10^{-19}\ \text{J}$ , so $\nu_{0A} = \dfrac{7.20\times10^{-19}}{6.63\times10^{-34}} = 1.09\times10^{15}\ \text{Hz}$ . For metal B: $\phi_{0B} = 2.1\times1.6\times10^{-19} = 3.36\times10^{-19}\ \text{J}$ , so $\nu_{0B} = \dfrac{3.36\times10^{-19}}{6.63\times10^{-34}} = 5.07\times10^{14}\ \text{Hz}$ .

Step 2. Compare each source frequency against each threshold. For metal A ( $\nu_{0A} = 1.09\times10^{15}\ \text{Hz}$ ): $\nu_1 = 6.0\times10^{14}\ \text{Hz}$ is below threshold, so no emission; $\nu_2 = 1.2\times10^{15}\ \text{Hz}$ exceeds threshold, so emission occurs. For metal B ( $\nu_{0B} = 5.07\times10^{14}\ \text{Hz}$ ): both $\nu_1$ and $\nu_2$ exceed threshold, so both frequencies cause emission.

Step 3. Find the ratio of threshold frequencies directly from the ratio of work functions, without recomputing joule values, since $\nu_0 \propto \phi_0$ :

\frac{\nu_{0A}}{\nu_{0B}} = \frac{\phi_{0A}}{\phi_{0B}} = \frac{4.5}{2.1} \approx 2.14

Step 4. State both final answers. From this source, only metal B emits electrons under $\nu_1$ , while both metals emit under $\nu_2$ ; and metal A's threshold frequency is about $2.14$ times metal B's. This is exactly the kind of "does it happen or not" reasoning competitive questions test, often layered on top of a numeric-answer part. Timing note: in a competitive exam, do not recompute both threshold frequencies from scratch if only a ratio is asked — spotting that $\nu_0 \propto \phi_0$ lets you skip straight to Step 3 and save the two joule-conversion calculations entirely; reserve the full Step 1 computation only for the parts that explicitly need a numeric frequency to compare against a given source.

Common traps

Treating the work function as a property of the light rather than the metal — $\phi_0$ never changes when you change the light source; only the photon energy $h\nu$ changes.
Believing a brighter (higher-intensity) beam can compensate for too-low a frequency — intensity controls the number of photons arriving per second, not the energy each one carries, so a dim high-frequency beam can emit electrons while an intense low-frequency beam emits none.
Forgetting the eV-to-joule conversion, or applying it backwards, before combining $\phi_0$ with $h$ (which is defined in joule-seconds) — this single unit slip is the most common way marks are lost on this topic.
Assuming every sample of a given metal has an identical work function — surface contamination, oxidation, or roughening genuinely changes $\phi_0$ , which is why textbook tables give typical, not universal, values.
Confusing threshold frequency $\nu_0$ (a fixed property, $\nu_0 = \phi_0/h$ ) with the frequency of light actually used in an experiment — the latter must be at or above $\nu_0$ for anything to happen at all.

You should be able to…

Explain, in terms of a surface energy barrier, why free electrons inside a metal do not normally leave on their own.
Define the work function $\phi_0$ , state its typical numerical range for common metals, and give it correctly in both eV and joules.
Name the three electron emission mechanisms — thermionic, field, and photoelectric — and describe in one sentence each how it supplies the required energy.
Derive and apply $\nu_0 = \phi_0/h$ to find a metal's threshold frequency, including a correct eV-to-joule conversion.
Explain why increasing the intensity of light below the threshold frequency still produces zero photoelectric emission.
Compare two metals' threshold frequencies directly from the ratio of their work functions, without recomputing each from scratch.

Ask the tutor: if you want to practise converting between work function, threshold frequency, and threshold wavelength for a few more metals before moving on to Einstein's photoelectric equation, ask the AI tutor to walk through extra examples with you.

Why do electrons not just fall back into the metal from wherever they came from, if they were free to move around inside it in the first place?

Inside the metal, an electron feels roughly equal pulls from ion cores in every direction, since it is surrounded by them on all sides, so it can wander with little net force. At the surface this symmetry breaks — there are ions behind but none in front — so the electron feels a net inward pull the moment it tries to leave. That net inward pull, integrated over the distance needed to escape, is exactly the work function energy it must overcome.

Is the work function the same thing as ionisation energy, since both sound like the energy needed to remove an electron?

They are related in spirit but not the same. Ionisation energy is the energy to remove an electron from a single isolated atom, leaving a charged ion behind. Work function is the energy to remove an electron from the surface of a bulk solid metal, where the electron was already a shared, delocalised conduction electron rather than bound to one atom. Work functions (typically a few eV) are usually smaller than the ionisation energies of the same element's isolated atoms.

Can visible light cause photoelectric emission from any metal, or does it depend on which metal it is?

It depends entirely on the metal's work function. Visible light photons carry roughly $1.8\ \text{eV}$ to $3.1\ \text{eV}$ of energy. Metals like caesium or potassium, with work functions around $2\ \text{eV}$ to $2.3\ \text{eV}$ , can be triggered by ordinary visible light. Metals like platinum, with a work function above $5\ \text{eV}$ , need ultraviolet light, since no visible photon carries enough energy to clear that barrier.

Why does this topic matter if the exam mostly asks about Einstein's photoelectric equation later in the chapter?

Because every quantity in Einstein's equation, $eV_0 = h\nu - \phi_0$ , is defined here first. If you are shaky on what $\phi_0$ physically means, why it is fixed for a metal, or how $\nu_0 = \phi_0/h$ works, the next article's algebra will feel like memorised symbol-shuffling instead of a story that makes sense — and board examiners regularly test the concept-check parts of this foundation directly, separate from the numerical equation itself.