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Physics · Dual Nature of Radiation and Matter

Einstein's Photoelectric Equation

In 1905 Einstein replaced the continuous-energy picture of light with discrete quanta and wrote a single energy-balance line that explained every feature of the photoelectric effect. This page follows NCERT Class 12 Physics §11.5–11.6 and NIOS §25.2 to build $K_{\max} = h\nu - \varphi_0$ from scratch, derive the threshold frequency and the stopping-potential line of slope $h/e$, and drill the result against real NEET questions where it is among the most consistently examined ideas in the chapter.

Why the wave theory fails

By the end of the nineteenth century the wave picture of light was firmly established: interference, diffraction and polarisation all followed naturally if light is an electromagnetic wave with energy spread continuously over the wavefront. NCERT §11.5 asks the decisive question — can this same wave picture account for the photoelectric effect? On three counts it cannot.

In the wave picture the energy a surface electron absorbs grows with the intensity, because intensity sets the amplitude of the electric and magnetic fields. The maximum kinetic energy of the photoelectrons should therefore rise with intensity, and a beam of any frequency, if intense enough and given enough time, should eventually free electrons. There would be no threshold frequency at all. Both predictions contradict experiment.

The wave picture also fails on timing. Energy is absorbed continuously across the whole wavefront, shared among an enormous number of electrons, so the energy delivered to one electron per unit time is tiny. Explicit estimates give hours before a single electron accumulates enough energy to escape — yet emission is observed to be instantaneous.

ObservationWave theory predictsExperiment shows
Effect of intensity on $K_{\max}$$K_{\max}$ increases with intensity$K_{\max}$ independent of intensity
Threshold frequencyNone — any frequency works if intense enoughSharp cut-off frequency $\nu_0$ exists
Time lag before emissionHours for a dim sourceInstantaneous ($\sim 10^{-9}\,$s or less)

The photon hypothesis

Einstein's resolution, set out in NCERT §11.6 and NIOS §25.2, was to abandon continuous absorption. Radiation energy is built up of discrete units — quanta — each carrying energy $E = h\nu$, where $h$ is Planck's constant and $\nu$ the frequency of the light. In the photoelectric process a single electron absorbs a single quantum. The whole quantum is either taken in or it is not; there is no gradual accumulation.

This single move reinterprets intensity. The intensity of monochromatic light is fixed by the number of quanta arriving per unit area per unit time, not by the energy of each quantum. Two beams of the same frequency but different intensity deliver photons of identical energy $h\nu$; the brighter beam simply delivers more of them per second.

Figure 1 · Energy budget Photon energy hν Work function φ₀ Kinetic energy K_max hν = φ₀ + K_max

The energy of one absorbed photon splits into a fixed escape cost $\varphi_0$ and whatever remains as kinetic energy. The most loosely bound electron keeps the maximum, $K_{\max}$.

The photoelectric equation

An electron at the surface must first pay the work function $\varphi_0$ — the minimum energy needed to escape the metal — before any energy is left over as motion. If the absorbed quantum exceeds this cost, the electron emerges. The most loosely bound electrons keep the largest share, so their maximum kinetic energy is

$$K_{\max} = h\nu - \varphi_0$$

This is Einstein's photoelectric equation (NCERT Eq. 11.2). More tightly bound electrons emerge with less than $K_{\max}$, because they lose extra energy escaping or in internal collisions. NIOS writes the same statement as $h\nu = \varphi_0 + K_{\max}$, a direct expression of energy conservation for the absorption of one photon by one electron.

NEET Trap

$K_{\max}$ tracks frequency, never intensity

A common error is to assume that brighter light gives faster photoelectrons. In Einstein's picture each electron absorbs exactly one photon of energy $h\nu$, so $K_{\max} = h\nu - \varphi_0$ is set by frequency alone. Doubling the intensity at the same frequency doubles the number of photons, doubling the photocurrent — but $K_{\max}$ and the stopping potential are unchanged.

Rule: Frequency controls $K_{\max}$ and $V_0$; intensity controls the number of photoelectrons (the current).

Threshold frequency

Kinetic energy cannot be negative, so emission is only possible when the photon energy exceeds the work function: $h\nu > \varphi_0$. The boundary case $K_{\max} = 0$ defines the smallest frequency that can free an electron. Setting $h\nu_0 = \varphi_0$ gives the threshold frequency (NCERT Eq. 11.3):

$$\nu_0 = \frac{\varphi_0}{h}$$

Below $\nu_0$ no photoelectrons are emitted no matter how intense the radiation or how long it falls on the surface, because no single photon carries enough energy. The threshold is a property of the metal: the greater the work function, the higher $\nu_0$. NIOS lists representative values — sodium $2.5\,$eV ($\nu_0 = 6.07\times10^{14}\,$Hz), potassium $2.3\,$eV ($5.58\times10^{14}\,$Hz), zinc $3.4\,$eV ($8.25\times10^{14}\,$Hz). Substituting $\varphi_0 = h\nu_0$ also lets the equation be written compactly as $K_{\max} = h(\nu - \nu_0)$.

Build the foundation

The stopping potential and threshold ideas come straight from the lab graphs. Revisit how they are measured in Experimental Study of the Photoelectric Effect.

Stopping potential and the slope h/e

The stopping potential $V_0$ is the negative collector voltage that just halts the most energetic photoelectron, so the work done against it equals the maximum kinetic energy: $K_{\max} = eV_0$ (NCERT Eq. 11.1; NIOS Eq. 25.1). Substituting into the photoelectric equation gives the experimentally testable form (NCERT Eq. 11.4):

$$eV_0 = h\nu - \varphi_0 \qquad\Longrightarrow\qquad V_0 = \frac{h}{e}\,\nu - \frac{\varphi_0}{e}$$

This is the equation of a straight line in the variables $V_0$ and $\nu$. Its slope is $h/e$ and its intercept on the frequency axis is the threshold $\nu_0$. Because $h$ and $e$ are universal constants, the slope is the same for every metal; only the intercept shifts. Millikan measured this slope for sodium and, using the known value of $e$, extracted Planck's constant $h$ — confirming Einstein's equation rather than disproving it as he had set out to do.

Figure 2 · V₀ versus ν ν V₀ ν₀ (A) Metal A ν₀ (B) Metal B slope = h / e (same for all metals)

Two metals give parallel lines of identical slope $h/e$. The line for the metal with the larger work function (B) is shifted right — its threshold frequency is higher and its intercept on the $V_0$ axis is more negative.

What the equation explains

The single line $K_{\max} = h\nu - \varphi_0$ resolves every observation that defeated the wave theory. NCERT §11.6 lays them out point by point, and they map directly onto the experimental laws.

ObservationEinstein's explanation
$K_{\max}$ depends on $\nu$, not intensityOne electron absorbs one quantum $h\nu$; intensity (number of quanta) is irrelevant to the single-electron process.
Threshold frequency $\nu_0$ exists$K_{\max}\ge 0$ requires $h\nu > \varphi_0$, i.e. $\nu > \nu_0 = \varphi_0/h$.
Photocurrent $\propto$ intensity (for $\nu>\nu_0$)More quanta per second means more electrons absorbing them, so more emission.
Emission is instantaneousAbsorption of one quantum by one electron is itself instantaneous; low intensity does not delay it.
NEET Trap

Below $\nu_0$, intensity buys nothing

If the incident frequency is below the threshold, no photoelectrons are emitted regardless of how intense or how prolonged the light is. A frequently set trap halves a frequency that was above threshold so that it falls below $\nu_0$, then doubles the intensity to suggest the current should grow. With $\nu < \nu_0$ the photocurrent is exactly zero — doubling the intensity of sub-threshold light produces no emission.

Rule: First check $\nu$ against $\nu_0$. If $\nu < \nu_0$, the current is zero before intensity is even considered.

Worked examples

The two NCERT examples below pin down the numerical use of the equation: the photon-counting interpretation of intensity, and the threshold-plus-stopping-potential calculation.

Example 1 · NCERT 11.1

Monochromatic light of frequency $6.0\times10^{14}\,$Hz from a laser emits power $2.0\times10^{-3}\,$W. Find (a) the energy of one photon and (b) the number of photons emitted per second.

(a) $E = h\nu = (6.63\times10^{-34})(6.0\times10^{14}) = 3.98\times10^{-19}\,$J.

(b) Power $P = NE$, so $N = P/E = (2.0\times10^{-3})/(3.98\times10^{-19}) = 5.0\times10^{15}$ photons per second. Intensity is literally a photon count rate.

Example 2 · NCERT 11.2

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

(a) $\nu_0 = \dfrac{\varphi_0}{h} = \dfrac{2.14\times1.6\times10^{-19}}{6.63\times10^{-34}} = 5.16\times10^{14}\,$Hz. Below this, no emission occurs.

(b) From $eV_0 = \dfrac{hc}{\lambda} - \varphi_0$, solve $\lambda = \dfrac{hc}{eV_0 + \varphi_0} = \dfrac{(6.63\times10^{-34})(3\times10^{8})}{(0.60 + 2.14)\times1.6\times10^{-19}} = 454\,$nm.

Quick Recap

Einstein's photoelectric equation in one screen

  • One electron absorbs one photon of energy $h\nu$: $K_{\max} = h\nu - \varphi_0$.
  • Threshold frequency $\nu_0 = \varphi_0/h$; below it no emission occurs at any intensity.
  • Stopping potential gives $K_{\max} = eV_0$, so $eV_0 = h\nu - \varphi_0$.
  • The $V_0$-versus-$\nu$ graph is a straight line of slope $h/e$, universal for all metals; only the intercept (threshold) changes.
  • Frequency sets $K_{\max}$ and $V_0$; intensity sets the photocurrent. Emission is instantaneous ($\sim10^{-9}\,$s).

NEET PYQ Snapshot — Einstein's Photoelectric Equation

Real NEET questions that turn on $K_{\max} = h\nu - \varphi_0$, threshold frequency and stopping potential.

NEET 2018

When light of frequency $2\nu_0$ ($\nu_0$ is the threshold frequency) is incident on a metal plate, the maximum velocity of electrons is $v_1$. When the frequency is increased to $5\nu_0$, the maximum velocity is $v_2$. The ratio $v_1 : v_2$ is

  1. $1 : 2$
  2. $1 : 4$
  3. $4 : 1$
  4. $2 : 1$
Answer: (1) 1 : 2

Using $\tfrac12 m v^2 = h\nu - h\nu_0$: for $2\nu_0$, $\tfrac12 m v_1^2 = h\nu_0$; for $5\nu_0$, $\tfrac12 m v_2^2 = 4h\nu_0$. Dividing, $v_1^2/v_2^2 = 1/4$, so $v_1 : v_2 = 1 : 2$.

NEET 2020

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What is the photoelectric current if the frequency is halved and the intensity is doubled?

  1. four times
  2. one-fourth
  3. zero
  4. doubled
Answer: (3) zero

The new frequency is $1.5\nu_0/2 = 0.75\nu_0$, which is below the threshold $\nu_0$. With $\nu < \nu_0$ no photoelectrons are emitted regardless of intensity, so the current is zero — the classic sub-threshold trap.

NEET 2023

The work functions of Cs, K and Na are $2.14\,$eV, $2.30\,$eV and $2.75\,$eV. If incident radiation has energy $2.20\,$eV, which surfaces may emit photoelectrons?

  1. Na only
  2. Cs only
  3. Both Na and K
  4. K only
Answer: (2) Cs only

Emission needs $h\nu \ge \varphi_0$. The photon energy $2.20\,$eV exceeds only the work function of caesium ($2.14\,$eV); K ($2.30\,$eV) and Na ($2.75\,$eV) need more energy than is supplied, so only Cs emits.

NEET 2016

A surface illuminated with radiation of wavelength $\lambda$ gives stopping potential $V$. With wavelength $2\lambda$ the stopping potential is $V/4$. The threshold wavelength for the surface is

  1. $5\lambda$
  2. $\lambda$
  3. $3\lambda$
  4. $4\lambda$
Answer: (3) 3λ

$eV = \tfrac{hc}{\lambda} - W$ and $\tfrac{eV}{4} = \tfrac{hc}{2\lambda} - W$. Eliminating $V$ gives $3W = \tfrac{hc}{\lambda}$. Since $W = \tfrac{hc}{\lambda_0}$, this yields $\lambda_0 = 3\lambda$.

FAQs — Einstein's Photoelectric Equation

The points examiners reword most often.

What is Einstein's photoelectric equation?
Einstein's photoelectric equation is K_max = hν − φ0, where K_max is the maximum kinetic energy of the emitted photoelectron, hν is the energy of the incident photon, and φ0 is the work function of the metal. It expresses energy conservation when a single electron absorbs a single quantum of radiation.
Why does the maximum kinetic energy not depend on the intensity of light?
In Einstein's picture each electron absorbs one photon of energy hν, and K_max = hν − φ0 is fixed by that single photon's frequency. Raising the intensity only increases the number of photons per second, so more electrons are emitted and the current rises, but the energy delivered to each electron, and therefore K_max, is unchanged.
What is the threshold frequency and how is it found from the equation?
The threshold frequency ν0 is the minimum frequency below which no photoelectrons are emitted, no matter how intense the light. Setting K_max = 0 in K_max = hν − φ0 gives hν0 = φ0, so ν0 = φ0/h. A larger work function means a higher threshold frequency.
Why is the slope of the V0-versus-ν graph the same for every metal?
Writing eV0 = hν − φ0 as V0 = (h/e)ν − φ0/e shows the graph of stopping potential against frequency is a straight line of slope h/e. Both h and e are universal constants, so the slope is the same for all metals. Only the intercept, which fixes the threshold frequency, changes from metal to metal.
How did Einstein's equation explain the instantaneous nature of emission?
The basic process is the absorption of one light quantum by one electron, which is instantaneous. As long as ν exceeds ν0, emission begins without any apparent time lag, of the order of 10^−9 s or less, even for very dim light. The wave theory, in which energy is absorbed continuously, predicted delays of hours and so failed this test.
How is Einstein's equation related to the stopping potential?
The stopping potential V0 is the negative collector voltage that just halts the most energetic photoelectron, so K_max = eV0. Substituting this into K_max = hν − φ0 gives eV0 = hν − φ0, which links the measured stopping potential directly to the photon frequency and the work function.
Related Topics
Dual Nature of Radiation and Matter Back to the full chapter overview and subtopic map. Experimental Study of Photoelectric Effect Saturation current, stopping potential and the V₀–ν graph in the lab. Photoelectric Effect: Hertz & Lenard The early observations the equation was built to explain. Particle Nature of Light: The Photon Photon energy hν and momentum hν/c that follow from the quanta idea. Electron Emission Work function and the three ways electrons leave a metal. Wave Nature of Matter: de Broglie The symmetry argument that gives particles a wavelength λ = h/p.