The Wave Theory on Trial

By 1900 the wave picture of light was triumphant — interference, diffraction and polarisation all explained naturally. In this picture, light is an electromagnetic wave with energy spread continuously over the wavefront. Let's see what that predicts for photoemission — and watch it fail, observation by observation.

Prediction 1: KmaxK_{max} should grow with intensity — it doesn't

Greater intensity means larger E and B field amplitudes, so each electron sitting in the wave should absorb more energy. The wave theory predicts KmaxK_{max} increases with intensity. Observed: KmaxK_{max} is completely independent of intensity.

Prediction 2: No threshold should exist — it does

Whatever the frequency, a sufficiently intense beam (acting for enough time) should eventually pump any electron past the work function. The wave theory forbids a threshold frequency. Observed: below ν0\nu_0, nothing — at any intensity.

Prediction 3: Dim light should mean long delays — it doesn't

In the wave picture the energy is shared across the entire wavefront, and huge numbers of electrons absorb it simultaneously, so each one collects energy at a trickle. Explicit calculation: a single electron would need hours or more to gather enough energy to escape. Observed: emission within 10−910^{-9} s, however dim the light.

Key Point: The wave picture contradicts observations (i), (ii) and (iii) of Section 4's summary, and its continuous-absorption assumption is demolished by the instantaneity observation (iv). It fails not narrowly but completely.

[NEET Important] Be able to list the three failures in one breath: intensity-independence of KmaxK_{max}, existence of threshold frequency, no time lag. 'Which observation can the wave theory explain?' — only that photocurrent grows with intensity.

Einstein's 1905 Idea: Light Comes in Packets

Albert Einstein's resolution was radical: photoelectric emission does not happen by continuous absorption. Instead,

Radiation energy is built of discrete units — quanta of energy hνh\nu — where h is Planck's constant and ν\nu the frequency of light.

The elementary process: one electron absorbs one quantum (hνh\nu). If this exceeds the minimum energy needed to escape (the work function ϕ0\phi_0), the electron comes out with maximum kinetic energy:

Kmax=hν−ϕ0K_{max} = h\nu - \phi_0

This is Einstein's photoelectric equation — Eq. (11.2). More tightly bound electrons emerge with less than KmaxK_{max} (they spend extra energy getting out), which explains the observed spread of electron energies.

How one line explains everything

  • KmaxK_{max} independent of intensity: the elementary act is one photon → one electron. Intensity = number of quanta arriving per unit area per unit time — irrelevant to the energy each electron receives. ✔
  • Threshold frequency: Kmax≥0K_{max} \geq 0 demands hν>ϕ0h\nu > \phi_0, i.e. ν>ν0\nu > \nu_0 where ν0=ϕ0h\nu_0 = \frac{\phi_0}{h} Greater work function → higher threshold. Below ν0\nu_0: no emission, however intense or prolonged the light. ✔
  • Photocurrent ∝ intensity: more quanta per second → more electrons absorb quanta → more electrons emitted per second (for ν>ν0\nu > \nu_0). ✔
  • Instantaneous emission: absorption of a quantum by an electron is a one-shot, instantaneous event. Low intensity means fewer events, never delayed events. ✔

Four observations, four ticks, one equation.

Einstein photoelectric equation and V0 versus frequency graph

The Straight Line That Measures Planck's Constant

Combine Kmax=eV0K_{max} = eV_0 (Section 4) with Einstein's equation:

eV0=hν−ϕ0(ν≥ν0)eV_0 = h\nu - \phi_0 \quad (\nu \geq \nu_0)

V0=heν−ϕ0e\boxed{V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}}

This predicts the V0V_0-vs-ν\nu graph is a straight line with:

  • slope = h/eh/e — a universal constant, independent of the material,
  • x-intercept = ν0=ϕ0/h\nu_0 = \phi_0/h (material-dependent),
  • y-intercept = −ϕ0/e-\phi_0/e (material-dependent).

Different metals give parallel lines — same slope, shifted thresholds. This is exactly what NCERT Fig. 11.5 shows.

Millikan: the sceptic who proved it

During 1906-1916, Millikan ran a decade of precision experiments aiming to disprove Einstein's equation. Instead, measuring the slope of the V0V_0-ν\nu line for sodium and using his own value of e, he extracted Planck's constant h — and it matched the value (6.626×10−346.626 \times 10^{-34} J s) from blackbody radiation, a completely different context. He verified the equation with great precision for several alkali metals over a wide frequency range. In 1916, Millikan had proved what he set out to demolish.

The hypothesis of light quanta, plus experimentally consistent values of h and ϕ0\phi_0, forced physics to accept Einstein's picture. Einstein received the 1921 Nobel Prize (citing the photoelectric effect); Millikan the 1923 Nobel Prize (elementary charge and photoelectric effect).

[JEE Tip] Graph questions cluster here: slope of V0V_0-ν\nu line = h/eh/e (same for all metals); the slope of the KmaxK_{max}-ν\nu line = h. From slope 4.12×10−154.12 \times 10^{-15} V s, get h=e×h = e \times slope =6.59×10−34= 6.59 \times 10^{-34} J s (that's NCERT Exercise 11.5). Also memorise the workhorse hc≈1240hc \approx 1240 eV nm for photon-energy conversions in one step.

Solved Examples

Example 1: Caesium end-to-end (NCERT Example 11.2)

The work function of caesium is 2.14 eV. Find (a) the threshold frequency, and (b) the wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.

Solution:

  1. (a) Threshold: ν0=ϕ0h=2.14×1.6×10−196.63×10−34\nu_0 = \dfrac{\phi_0}{h} = \dfrac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}.
  2. Calculate: ν0=5.16×1014\nu_0 = 5.16 \times 10^{14} Hz. Below this, no photoelectrons.
  3. (b) Einstein's equation with eV0eV_0: eV0=hcλ−ϕ0⇒λ=hceV0+ϕ0eV_0 = \dfrac{hc}{\lambda} - \phi_0 \Rightarrow \lambda = \dfrac{hc}{eV_0 + \phi_0}.
  4. Substitute: λ=(6.63×10−34)(3×108)(0.60+2.14)×1.6×10−19=19.89×10−262.74×1.6×10−19\lambda = \dfrac{(6.63 \times 10^{-34})(3 \times 10^8)}{(0.60 + 2.14) \times 1.6 \times 10^{-19}} = \dfrac{19.89 \times 10^{-26}}{2.74 \times 1.6 \times 10^{-19}}.
  5. Answer: λ≈454\lambda \approx 454 nm.
  6. Takeaway: eV0+ϕ0eV_0 + \phi_0 = photon energy — build the photon first, then convert to wavelength.

Example 2: The quick version with hc = 1240 eV nm [JEE Shortcut]

Redo Example 1(b) using hc=1240hc = 1240 eV nm.

Solution:

  1. Photon energy needed: E=eV0+ϕ0=0.60+2.14=2.74E = eV_0 + \phi_0 = 0.60 + 2.14 = 2.74 eV.
  2. Wavelength: λ=12402.74≈452\lambda = \dfrac{1240}{2.74} \approx 452 nm.
  3. Takeaway: the 1240 shortcut turns a five-line SI computation into one division (small rounding differences are expected and fine for MCQs).

Example 3: Maximum kinetic energy and speed (NCERT Exercise 11.2 style)

Caesium (ϕ0\phi_0 = 2.14 eV) is illuminated by light of frequency 6×10146 \times 10^{14} Hz. Find (a) KmaxK_{max}, (b) the stopping potential, (c) the maximum speed of photoelectrons.

Solution:

  1. Photon energy: hν=(6.63×10−34)(6×1014)1.6×10−19≈2.49h\nu = \dfrac{(6.63 \times 10^{-34})(6 \times 10^{14})}{1.6 \times 10^{-19}} \approx 2.49 eV.
  2. (a) Kmax=hν−ϕ0=2.49−2.14=0.35K_{max} = h\nu - \phi_0 = 2.49 - 2.14 = 0.35 eV.
  3. (b) V0=Kmax/e=0.35V_0 = K_{max}/e = 0.35 V.
  4. (c) vmax=2Kmaxm=2×0.35×1.6×10−199.1×10−31≈3.5×105v_{max} = \sqrt{\dfrac{2K_{max}}{m}} = \sqrt{\dfrac{2 \times 0.35 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 3.5 \times 10^{5} m/s.
  5. Takeaway: the chain hν→Kmax→V0→vmaxh\nu \to K_{max} \to V_0 \to v_{max} is the chapter's standard four-step ladder.

Example 4: Predicting a cut-off voltage (NCERT Exercise 11.6)

The threshold frequency for a metal is 3.3×10143.3 \times 10^{14} Hz. Light of frequency 8.2×10148.2 \times 10^{14} Hz is incident. Predict the cut-off voltage.

Solution:

  1. Formula: eV0=h(ν−ν0)eV_0 = h(\nu - \nu_0).
  2. Substitute: V0=6.63×10−34×(8.2−3.3)×10141.6×10−19V_0 = \dfrac{6.63 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}}.
  3. Calculate: V0=6.63×4.9×10−201.6×10−19≈2.0V_0 = \dfrac{6.63 \times 4.9 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 2.0 V.
  4. Takeaway: when ν0\nu_0 is given, skip ϕ0\phi_0 entirely — work with h(ν−ν0)h(\nu - \nu_0) directly.

Example 5: Will it emit? (NCERT Exercise 11.7)

The work function of a metal is 4.2 eV. Will it emit photoelectrons for radiation of wavelength 330 nm?

Solution:

  1. Photon energy: E=hcλ=1240 eV nm330 nm≈3.76E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{330 \text{ nm}} \approx 3.76 eV.
  2. Compare: 3.76 eV < 4.2 eV = ϕ0\phi_0.
  3. Answer: No emission — the photon falls short of the work function, and no amount of intensity changes that.
  4. Takeaway: always settle the yes/no question (photon vs ϕ0\phi_0) before computing anything else.

Example 6: Planck's constant from a slope (NCERT Exercise 11.5)

The slope of the cut-off voltage versus frequency graph is 4.12×10−154.12 \times 10^{-15} V s. Find h.

Solution:

  1. Theory: V0=heν−ϕ0eV_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}, so slope = h/eh/e.
  2. Rearrange: h=e×h = e \times slope =1.6×10−19×4.12×10−15= 1.6 \times 10^{-19} \times 4.12 \times 10^{-15}.
  3. Answer: h=6.59×10−34h = 6.59 \times 10^{-34} J s.
  4. Takeaway: this is literally Millikan's method — the slope of that line measured Planck's constant from photoelectricity.

Example 7: Threshold from photoelectron speed (NCERT Exercise 11.8)

Light of frequency 7.21×10147.21 \times 10^{14} Hz is incident on a metal; electrons with maximum speed 6.0×1056.0 \times 10^5 m/s are ejected. Find the threshold frequency.

Solution:

  1. Maximum kinetic energy: Kmax=12mv2=12(9.1×10−31)(6.0×105)2=1.64×10−19K_{max} = \frac{1}{2}mv^2 = \frac{1}{2}(9.1 \times 10^{-31})(6.0 \times 10^5)^2 = 1.64 \times 10^{-19} J.
  2. Einstein's equation: hν0=hν−Kmaxh\nu_0 = h\nu - K_{max}.
  3. Substitute: ν0=ν−Kmaxh=7.21×1014−1.64×10−196.63×10−34\nu_0 = \nu - \dfrac{K_{max}}{h} = 7.21 \times 10^{14} - \dfrac{1.64 \times 10^{-19}}{6.63 \times 10^{-34}}.
  4. Calculate: ν0=7.21×1014−2.47×1014=4.74×1014\nu_0 = 7.21 \times 10^{14} - 2.47 \times 10^{14} = 4.74 \times 10^{14} Hz.
  5. Takeaway: speed → energy → subtract from photon frequency: a three-step backward pass through Einstein's equation.

Example 8: Work function from an argon laser (NCERT Exercise 11.9)

Light of wavelength 488 nm from an argon laser gives photoelectrons with stopping potential 0.38 V. Find the emitter's work function.

Solution:

  1. Photon energy: E=1240488≈2.54E = \dfrac{1240}{488} \approx 2.54 eV.
  2. Einstein rearranged: ϕ0=E−eV0=2.54−0.38\phi_0 = E - eV_0 = 2.54 - 0.38.
  3. Answer: ϕ0≈2.16\phi_0 \approx 2.16 eV.
  4. Takeaway: photon energy minus stopping energy = work function; every quantity in eV keeps it one clean subtraction.

Example 9: Two wavelengths, one metal [JEE Multi-step]

For a metal, stopping potentials are 1.0 V with 400 nm light and 0.22 V with 500 nm light (illustrative values). Show how h can be extracted from just these two readings.

Solution:

  1. Write the equation twice: eV01=hcλ1−ϕ0eV_{01} = \dfrac{hc}{\lambda_1} - \phi_0 and eV02=hcλ2−ϕ0eV_{02} = \dfrac{hc}{\lambda_2} - \phi_0.
  2. Subtract (kills ϕ0\phi_0): e(V01−V02)=hc(1λ1−1λ2)e(V_{01} - V_{02}) = hc\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right).
  3. Solve for h: h=e(V01−V02)c(1λ1−1λ2)=1.6×10−19×0.783×108×(2.5−2.0)×106h = \dfrac{e(V_{01} - V_{02})}{c\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)} = \dfrac{1.6 \times 10^{-19} \times 0.78}{3 \times 10^8 \times (2.5 - 2.0) \times 10^{6}}.
  4. Calculate: h≈8.3×10−34h \approx 8.3 \times 10^{-34} J s — order-correct for these rounded illustrative readings; precise data give 6.6×10−346.6 \times 10^{-34} J s.
  5. Takeaway: the two-wavelength subtraction eliminates the unknown work function — a standard JEE construction.

Example 10: Why the wave theory predicts hours [Board Conceptual]

Outline the wave-theory time-lag estimate and why it fails.

Solution:

  1. In the wave picture, energy arrives continuously spread over the whole wavefront, shared among an enormous number of surface electrons.
  2. The power collected by the tiny cross-section of a single electron is minuscule; explicit calculation shows hours or more are needed to accumulate a few eV.
  3. Experiment: emission within 10−910^{-9} s at any intensity.
  4. Conclusion: continuous absorption is wrong; energy must arrive in concentrated packets (hνh\nu) absorbed whole by single electrons — Einstein's elementary process.