NDA Physics · Modern Physics

Photoelectric Effect: Light as Particles

When light of high enough frequency strikes a metal surface, it ejects electrons instantly; the energy of each ejected electron depends on the light's frequency (colour), not its brightness.

All Modern Physics notes NDA Physics strategy

Why this matters

This is the gateway to all of modern physics: it is the experiment that forced light to be treated as particles (photons), each carrying energy E = hf. The NDA tests it three ways — the name of the phenomenon, who explained it (Einstein, who won the Nobel for exactly this), and one-step plug-ins using E = hf or the X-ray cutoff wavelength. Four PYQs, all EASY or MODERATE.

Concept 1 of 4

The photon — light carries energy in discrete packets E = hf

Intuition

Before modern physics, light was purely a wave. The photoelectric effect showed light also behaves as a stream of particles called photons. Each photon carries a fixed packet of energy that depends only on the light's frequency: high-frequency light (violet, ultraviolet, X-rays) has high-energy photons; low-frequency light (red, infrared) has low-energy photons. Brightness just means MORE photons, not more energetic ones.

Definition

Light is made of photons, each carrying energy:

Energy per photon: $E = hf = \dfrac{hc}{\lambda}$ , where $f$ is frequency, $\lambda$ is wavelength, $c$ is the speed of light.
Planck's constant $h \approx 6.63 \times 10^{-34}$ J·s — the dimensions of $h$ are those of angular momentum (J·s = energy x time).
Higher frequency = shorter wavelength = more energy per photon. So among radio, light, and X-rays, X-rays carry the most energy per photon.
This particle picture of light is its dual nature — light is both wave and particle.

Energy of a photon

E = hf = \dfrac{hc}{\lambda}

Eenergy of one photon (J)
hPlanck's constant, 6.63 x 10⁻³⁴ J·s
ffrequency of the light (Hz)
cspeed of light, 3 x 10⁸ m/s
λwavelength of the light (m)

Light above the threshold frequency ejects electrons instantly; they cross to the anode and a current flows. Below threshold, no electrons emerge no matter how bright the light.

Worked example

Among radio waves, visible light, and X-rays, which carries the maximum energy per photon?

Energy per photon is $E = hf$ — it grows with frequency.
Order by frequency: radio waves (lowest) < visible light < X-rays (highest).
Therefore X-rays, having the highest frequency, carry the most energy per photon.

Answer:X-rays carry the maximum energy per photon.

Practice this conceptself-check · 4 quick reps

Try it yourself

Photon A has twice the frequency of photon B. How do their energies compare?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What is the energy of a single photon equal to?
2.
Which has the most energy per photon: infrared, visible light, or ultraviolet?
3.
The dimensions of Planck's constant h are the same as which quantity?
4.
If wavelength increases, does photon energy increase or decrease?

From the bank · past-year question

Example 1Modern PhysicsEASY

Which one among the following waves carries the maximum energy per photon?

[Q114 · Sep · 2017]

Brightness does NOT change photon energy

Making light brighter sends MORE photons per second, but each photon still carries the same energy

E = hf

. To raise the energy per photon you must raise the frequency (shift toward blue/UV/X-ray), not turn up the intensity.

Energy grows with frequency, falls with wavelength

E = hf = hc/\lambda

. Since

f

and

\lambda

are inversely related, an option saying "longer wavelength = more energy" is always wrong. Long wavelength (red, radio) = low energy.

Concept 2 of 4

Photoelectric emission — light ejecting electrons from a metal

Intuition

Shine light on a clean metal surface and, if the light's frequency is high enough, electrons are knocked out of the metal almost instantly. These ejected electrons are called photoelectrons, and the whole phenomenon is photoelectric emission. There is a minimum frequency (the threshold) below which NO electrons come out, however bright the light.

Definition

Photoelectric emission is the ejection of electrons from a metal surface when light of a sufficiently high frequency falls on it.

The emitted electrons are called photoelectrons.
There is a threshold frequency for each metal — below it, no emission occurs no matter how intense the light.
Above threshold, emission is instantaneous and the number of photoelectrons rises with the light's intensity (brightness).

Worked example

What is the emission of electrons from a metallic surface by the application of light called?

Light supplies photons to the metal surface.
When a photon's energy exceeds the metal's threshold, an electron is ejected.
This process — electron ejection by light — is named photoelectric emission.

Answer:Photoelectric emission (or the photoelectric effect).

Practice this conceptself-check · 4 quick reps

Try it yourself

A metal does not emit electrons under bright red light but emits them under dim violet light. Why?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What are the electrons ejected by light called?
2.
Does increasing the brightness of below-threshold light cause emission?
3.
Emission of electrons from a metal by light is called what?
4.
Above the threshold, what does raising intensity increase?

From the bank · past-year question

Example 2Modern PhysicsEASY

Electron emission from a metallic surface by application of light is known as

[Q73 · Sep · 2017]

Threshold is about FREQUENCY, not intensity

A bright low-frequency beam ejects zero electrons; a faint high-frequency beam ejects them at once. The deciding factor is whether each photon clears the threshold frequency — never the brightness.

Concept 3 of 4

Who explained the photoelectric effect — Einstein and the Nobel Prize

Intuition

The experiment was a puzzle for classical physics, which predicted brighter light should always eject electrons. Albert Einstein explained it in 1905 using the photon idea: one photon gives all its energy to one electron, so frequency (photon energy), not brightness, decides emission. He won the 1921 Nobel Prize in Physics for this — not for relativity. This is a favourite recall question.

Definition

Albert Einstein explained the photoelectric effect (1905) by treating light as photons, each delivering energy $hf$ to a single electron. He received the 1921 Nobel Prize in Physics for this explanation.

Person / idea	Contribution
Albert Einstein	Explained the photoelectric effect using the photon/quantum idea (1905) NDA 2019 — the photoelectric effect was explained by Albert Einstein (not Bohr, Planck, or Rutherford).
Max Planck	Introduced energy quanta $E = hf$ (Planck's constant); the quantum seed Einstein used
Heinrich Hertz	First OBSERVED the photoelectric effect experimentally (but did not explain it)

Distinguish the OBSERVER (Hertz) from the EXPLAINER (Einstein). The exam asks who explained it — that is Einstein.

Practice this conceptself-check · 4 quick reps

Try it yourself

Who among Bohr, Einstein, Rutherford, and Newton explained the photoelectric effect, and what award did it earn?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Who explained the phenomenon of the photoelectric effect?
2.
For which work did Einstein receive his Nobel Prize?
3.
Who first experimentally observed the photoelectric effect?
4.
Whose energy-quantum idea (E = hf) did Einstein build on?

From the bank · past-year question

Example 3Modern PhysicsEASY

Who among the following has explained the phenomenon of photoelectric effect?

[Q85 · Apr · 2019]

Einstein's Nobel was for the photoelectric effect, NOT relativity

A classic distractor pairs Einstein's Nobel Prize with relativity. The 1921 Nobel Prize was awarded specifically for his explanation of the photoelectric effect.

Observed vs explained

Hertz observed the effect; Einstein explained it. The exam wording "explained the phenomenon" points to Einstein.

Concept 4 of 4

Cutoff wavelength and the energy-voltage link

Intuition

In an X-ray tube, electrons are accelerated through a voltage V, gaining energy eV, and then smash into a target to make X-rays. The most energetic X-ray photon possible carries all of that energy, which fixes the SHORTEST wavelength produced — the cutoff wavelength. Because energy and wavelength are inversely related, raising the voltage shortens the cutoff wavelength.

Definition

The cutoff (minimum) wavelength of X-rays from a tube run at accelerating voltage $V$ is set by equating the electron's energy to the maximum photon energy:

$eV = \dfrac{hc}{\lambda_{min}}$ , so $\lambda_{min} = \dfrac{hc}{eV}$ .
$\lambda_{min}$ is inversely proportional to V — double the voltage and the cutoff wavelength is halved.

Cutoff wavelength of X-rays

\lambda_{min} = \dfrac{hc}{eV}

λ_minshortest (cutoff) wavelength produced
hPlanck's constant
cspeed of light
eelectron charge
Vaccelerating voltage of the tube

Worked example

The voltage applied to an X-ray tube is doubled, with the filament-target separation unchanged. What happens to the cutoff wavelength?

Cutoff wavelength is $\lambda_{min} = \dfrac{hc}{eV}$ — inversely proportional to $V$ .
Doubling $V$ multiplies the denominator by 2.
So $\lambda_{min}$ becomes half of its original value.

Answer:The cutoff wavelength is halved.

Practice this conceptself-check · 4 quick reps

Try it yourself

If the X-ray tube voltage is reduced to one-third, what happens to the cutoff wavelength?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Cutoff wavelength is proportional to what power of the tube voltage?
2.
Double the tube voltage: cutoff wavelength becomes?
3.
Halve the tube voltage: cutoff wavelength becomes?
4.
What energy does an electron gain crossing a voltage V?

From the bank · past-year question

Example 4Modern PhysicsMODERATE

If the potential difference applied to an X-ray tube is doubled while keeping the separation between the filament and the target as same, what will happen to the cutoff wavelength ?

[Q122 · Apr · 2017]

Cutoff wavelength is inversely related to voltage

Double V does NOT double the wavelength — it halves it. The energy goes up, and higher energy means shorter wavelength. Distractors offering "doubled" or "four times" reverse the relationship.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The photon — light carries energy in discrete packets E = hf
Energy of a photon
$E = hf = \dfrac{hc}{\lambda}$
Cutoff wavelength and the energy-voltage link
Cutoff wavelength of X-rays
$\lambda_{min} = \dfrac{hc}{eV}$

Reference tables (1)

Who explained the photoelectric effect — Einstein and the Nobel Prize3 rows

Person / idea	Contribution
Albert Einstein	Explained the photoelectric effect using the photon/quantum idea (1905) NDA 2019 — the photoelectric effect was explained by Albert Einstein (not Bohr, Planck, or Rutherford).
Max Planck	Introduced energy quanta $E = hf$ (Planck's constant); the quantum seed Einstein used
Heinrich Hertz	First OBSERVED the photoelectric effect experimentally (but did not explain it)

Distinguish the OBSERVER (Hertz) from the EXPLAINER (Einstein). The exam asks who explained it — that is Einstein.

Watch out for (6)

Brightness does NOT change photon energy
→ The photon — light carries energy in discrete packets E = hf
Energy grows with frequency, falls with wavelength
→ The photon — light carries energy in discrete packets E = hf
Threshold is about FREQUENCY, not intensity
→ Photoelectric emission — light ejecting electrons from a metal
Einstein's Nobel was for the photoelectric effect, NOT relativity
→ Who explained the photoelectric effect — Einstein and the Nobel Prize
Observed vs explained
→ Who explained the photoelectric effect — Einstein and the Nobel Prize
Cutoff wavelength is inversely related to voltage
→ Cutoff wavelength and the energy-voltage link

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