Einstein’s photoelectric equation

What is Einstein’s photoelectric equation?

In plain words: light pays electrons in fixed coins, one coin each, and each coin is worth hν. An electron must pay an exit fee, the work function, and keeps the change as kinetic energy. Brighter light means more coins, so more electrons, but not bigger coins; only a higher frequency makes each coin worth more.

Regime: the standard one-photon photoelectric model at ordinary intensity, with a fixed clean surface and negligible heating, space charge, contact-potential offsets and background current (reasoned). “Instantaneous” means no classical energy-accumulation delay, not literally zero interaction time or zero random waiting time for the next photon (reasoned). Beyond the NCERT text: not every photon produces an emitted electron; the chance depends on the photon energy (Wikipedia, “Photoelectric effect”). Electron speeds from ½mv² are nonrelativistic. The linear stopping-potential relation applies above threshold; extending it below threshold does not describe emitted electrons with negative kinetic energy (reasoned).

Energy quantum hν

• In Einstein’s picture, emission does not come from energy absorbed continuously. Each quantum of radiant energy has energy hν, and in the photoelectric effect an electron absorbs one quantum (NCERT page 281).
• The intensity of light of a given frequency is set by the number of quanta (photons) arriving per unit area per second; the energy of each quantum depends only on the frequency (NCERT pages 281–282).
• Planck’s constant is about 6.626 × 10⁻³⁴ J s (NCERT page 283). In electron volts, E = hν gives about 2.5 eV for light of 6 × 10¹⁴ Hz (reasoned; see Example 1).

K_max = hν − φ_₀

• If the quantum’s energy exceeds the work function, the electron is emitted, and the most loosely bound electrons leave with the maximum kinetic energy (NCERT Eq. (11.2), page 281):

• More tightly bound electrons come out with less than K_max (NCERT page 281).
• K_max depends linearly on ν and not at all on the intensity, because the effect comes from one quantum absorbed by one electron. More intense light gives more quanta, so more electrons and a larger current, but each still gets hν (NCERT page 282).
• The basic process is the absorption of a light quantum by an electron, a single event, so photoelectric emission is instantaneous however dim the light (NCERT page 282); NCERT puts the delay at about 10⁻⁹ s or less (page 280). Strictly, “instantaneous” means prompt on that timescale, not a zero-duration process (reasoned).

Threshold frequency ν_₀ = φ_₀/h

• Since K_max cannot be negative, photoelectric emission is possible only if hν > φ₀, that is ν > ν₀ (NCERT Eq. (11.3), page 282); at the threshold itself, hν = φ₀ and K_max = 0 (reasoned):

• The larger the work function, the higher the threshold frequency. Below ν₀ no electrons come out, however intense the light or however long it shines (NCERT page 282).
• In terms of ν₀, K_max = h(ν − ν₀) (NCERT Summary point 7, page 287).
• In terms of wavelength, emission needs λ < λ₀, where λ₀ is the threshold wavelength (reasoned from NCERT Eq. (11.3) and c = νλ):

• For caesium, λ₀ = (3 × 10⁸)/(5.16 × 10¹⁴) ≈ 581 nm (see Example 2): violet light ejects electrons from it, red light does not (reasoned).

Stopping potential against frequency: slope h/e

• With K_max = eV₀ (the energy an electron loses against a retarding potential V₀), the equation becomes eV₀ = hν − φ₀, so (NCERT Eq. (11.4), page 282):

• This holds for ν ≥ ν₀. The graph of V₀ against ν is a straight line of slope h/e, the same for every material; it meets the frequency axis at ν₀ (NCERT Figure 11.5, page 279; Eq. (11.4), pages 282–283). Produced backwards, it would meet the V₀ axis at −φ₀/e. Plotted as K_max against ν instead, the line has slope h and meets the K_max axis at −φ₀ (both reasoned).
• Millikan’s test. From 1906 to 1916 Millikan did experiments aimed at disproving the equation. He measured the slope of the line for sodium and, with the known e, found h close to 6.626 × 10⁻³⁴ J s, the value found in an entirely different context. In 1916 he had proved the equation instead of disproving it (NCERT page 283).
• Einstein received the 1921 Nobel Prize for his contribution to theoretical physics and the photoelectric effect; Millikan received the 1923 prize, partly for his work on the photoelectric effect (NCERT page 283).

Solved examples

Common mistakes to avoid

Quick revision summary

• Light energy comes in quanta of E = hν; one electron absorbs one quantum.
• K_max = hν − φ₀: linear in frequency, independent of intensity.
• Threshold frequency ν₀ = φ₀/h; no emission below it; K_max = h(ν − ν₀).
• With K_max = eV₀: V₀ = (h/e)ν − φ₀/e, a straight line of slope h/e for every metal.
• Millikan’s measurements (1906–1916) confirmed the equation and gave h ≈ 6.6 × 10⁻³⁴ J s.

Questions students ask

Can Einstein’s photoelectric equation be derived, or is it just conservation of energy?

Einstein’s photoelectric equation is the law of conservation of energy applied to one quantum of light absorbed by one electron (NCERT Summary point 7, page 287), so its ‘derivation’ is a single energy balance, hν = φ₀ + K_max. It gives the maximum kinetic energy because the work function is the least energy needed to escape: an electron that needs more than φ₀ keeps less (NCERT page 281). The real step was Einstein’s hypothesis that light energy comes in whole quanta of hν, each absorbed by one electron (NCERT page 281); measuring K_max by the stopping potential then turns the same balance into NCERT Eq. (11.4).

What is the difference between threshold frequency and work function?

The work function φ₀ is an energy, the least an electron needs to escape from the metal, while the threshold frequency ν₀ is the frequency of light whose photons carry exactly that energy, so φ₀ = hν₀ (NCERT Eq. (11.3), page 282). Both describe the same property of the metal surface: dividing a work function in eV by h ≈ 4.14 × 10^⁻¹⁵ eV s gives the threshold frequency, so a metal of work function 3.6 eV has ν₀ ≈ 8.7 × 10^¹⁴ Hz, in the ultraviolet. The work function is measured in joules or electron volts, [M L^² T^⁻²], and the threshold frequency in hertz, [T^⁻¹] (NCERT page 288).

Are threshold frequency and cut-off frequency the same thing?

Yes, in the photoelectric effect the threshold frequency and the cut-off frequency are the same: NCERT first calls ν₀ the minimum cut-off frequency, at which the stopping potential falls to zero, then names it the threshold frequency (pages 279–280). Below it no photoelectrons come out, however intense the light. For X-rays the cut-off is instead the highest frequency a tube gives, ν_max = eV/h (beyond the NCERT text; energy conservation for an electron stopped in an X-ray target), as in NCERT Exercise 11.1.

If the frequency of light is doubled, does the maximum kinetic energy of photoelectrons double?

For an initial frequency already above threshold, doubling the frequency more than doubles the maximum kinetic energy, because the work function is taken off only once: K_max = hν − φ₀ becomes 2hν − φ₀ = 2K_max + φ₀ (from NCERT Eq. (11.2), page 281). For example, photons of 3.2 eV on a metal of work function 2.4 eV give K_max = 0.8 eV, but photons of 6.4 eV give 4.0 eV, five times as much, and the maximum speed rises √5 ≈ 2.2 times. K_max and the stopping potential rise in a straight line with ν but are not proportional to it, because the line starts at the threshold ν₀, not at zero.

How do you find the threshold frequency without knowing the work function?

Without the work function, the threshold frequency comes from the stopping potential: if h is known, one V₀ at one frequency gives ν₀ = ν − eV₀/h (from NCERT Eq. (11.4), page 282). If h is not known either, measure V₀ at two frequencies: the line through the two points has slope h/e and cuts the frequency axis at ν₀ (NCERT Figure 11.5, page 279). For 0.45 V at 6.5 × 10^¹⁴ Hz and 1.70 V at 9.5 × 10^¹⁴ Hz, the slope is 1.25 V/(3.0 × 10^¹⁴ Hz) ≈ 4.17 × 10^⁻¹⁵ V s, so 0.45 V of the line spans 1.08 × 10^¹⁴ Hz and ν₀ ≈ 6.5 × 10^¹⁴ − 1.08 × 10^¹⁴ ≈ 5.4 × 10^¹⁴ Hz.

Why is there no time lag in the photoelectric effect?

There is no classical energy-accumulation lag: a successful single-photon absorption can supply the escape energy at once, so, as NCERT puts it, “Low intensity does not mean delay in emission” (page 282). Strictly, real absorption and transport take finite times, and a weak beam can leave a long random wait before an individual photon is absorbed (reasoned). Dim light only brings fewer photons each second, so fewer electrons and a smaller current, without reducing the maximum energy of each emitted electron: even a beam of 1.0 × 10^⁻¹² W of 470 nm light delivers about 2.4 × 10^⁶ photons every second. Experiment finds emission within about 10^⁻⁹ s or less (NCERT page 280).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Will a metal of work function 4.2 eV emit photoelectrons in light of wavelength 330 nm? (NCERT Exercise 11.7)

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   hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(330 × 10⁻⁹) ≈ 6.0 × 10⁻¹⁹ J ≈ 3.8 eV, less than 4.2 eV, so no (NCERT Exercise 11.7, page 289; answer on page 351).

2. Q2. A metal has a work function of 2.0 eV. Find its threshold frequency.

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   ν₀ = (2.0 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) ≈ 4.8 × 10¹⁴ Hz (NCERT Eq. (11.3), page 282).

3. Q3. Light of energy 5.0 eV falls on a metal of work function 3.0 eV. Find K_max and the stopping potential.

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   K_max = 5.0 − 3.0 = 2.0 eV, so V₀ = 2.0 V (NCERT Eqs. (11.2) and (11.4), pages 281–282).