Q.The work function of caesium metal is 2.14 eV. When light of frequency 6×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
Concept understanding — Photon Energy
Photon Energy: The Energy Carried by Light
Light does not carry its energy in a continuous stream. It comes in discrete packets — like individual drops instead of a steady hose. Each packet is called a photon: the smallest possible unit of light of a given frequency. You cannot have half a photon; it is all or nothing.
The Intuition: Colour Determines Energy
Red light and blue light are both light, but they behave differently — blue light causes sunburn and drives chemical reactions more readily than red. The reason is that the colour (frequency) of light directly fixes how much energy each photon carries, and blue photons carry more energy than red photons.
The Planck–Einstein Relation
The energy E of a single photon is directly proportional to its frequency f, and inversely proportional to its wavelength λ:
E=hf=λhc
Where:
- E = energy of one photon (joules, J)
- f = frequency (hertz, Hz)
- λ = wavelength (metres, m)
- c = speed of light in vacuum (3.00×108 m/s)
- h = Planck's constant (6.626×10−34 J·s)
Planck's constant h is extremely tiny, so an individual photon carries a very small amount of energy — which is why we never feel single photons striking our skin.
What This Tells You
- Higher frequency = higher energy. Gamma rays have extremely energetic photons; radio waves have very low-energy photons.
- Shorter wavelength = higher energy. Ultraviolet photons are more energetic than visible-light photons; infrared photons are less.
- Energy is quantised. Light energy comes in fixed packets — you can have 1, 2, or 1000 photons, but never 0.5. This was Max Planck's revolutionary idea of 1900.
This is why UV light causes sunburn (UV photons carry enough energy to damage DNA, while visible photons do not), why X-ray photons are energetic enough to pass through soft tissue but get absorbed by bone, and why a photon needs enough energy to actually break a bond or drive a photochemical/biological reaction, rather than merely wavelength-dependent intensity.
A Concrete Example
Question: Which has more energy — a photon of red light (λ=700 nm) or a photon of blue light (λ=450 nm)?
Since E=λhc, energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:
- Red: Ered=700×10−9(6.626×10−34)(3.00×108)≈2.84×10−19 J
- Blue: Eblue=450×10−9(6.626×10−34)(3.00×108)≈4.42×10−19 J
The blue photon carries about 1.5 times the energy of the red photon.
A common mistake is to think brighter light means more energetic photons. Brightness is the number of photons per second, not the energy per photon. A bright red light has many low-energy photons; a dim blue light has fewer but higher-energy ones.
The Big Picture
Photon energy is the bridge between the wave nature of light (frequency, wavelength) and its particle nature (energy packets) — one of the foundational ideas of quantum mechanics: at the smallest scales, energy is not continuous but comes in discrete, indivisible units.
Photon energy, given by the Planck-Einstein relation E = hf, is one of the most fundamental formulas in the NCERT Class 12 Physics Dual Nature of Radiation and Matter chapter, and "photon energy formula and calculation" is a heavily searched query among students preparing for CBSE boards, JEE Main, and NEET. Because this idea links directly to the photoelectric effect and atomic spectra, it also anchors several "modern physics important questions" compiled for competitive-exam revision.
Why this formula?
Why Photon Energy is E=hf — The Reasoning Behind the Formula
The formula E=hf is not something you memorise and plug numbers into. It comes from a deep shift in how physicists understood light.
The problem that forced the formula
By the late 1800s, classical physics said light was a wave — continuous, carrying energy in proportion to its intensity (brightness). But blackbody radiation and the photoelectric effect showed something bizarre: when you shine light on a metal, electrons are ejected only if the light's frequency is above a certain threshold, no matter how bright the light is. Below that frequency, even the brightest lamp won't kick out a single electron. Classically this made no sense — a wave's energy depends on amplitude, not frequency.
Einstein's radical idea (1905)
Einstein proposed that light comes in discrete packets — photons — each carrying a fixed energy that depends only on its frequency, with Planck's constant h as the proportionality:
A single photon's energy is E=hf, where f is the frequency of the light.
This directly explains the photoelectric effect: an electron needs a minimum energy (the work function ϕ) to escape. If a photon's energy hf<ϕ, no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.
E=hf is not derived from deeper principles — it is a postulate justified by the experiments it explains, and it unified optics with quantum mechanics.
The equivalent forms
Since c=fλ for light in vacuum, the same idea in terms of wavelength is:
E=λhc
Use this when you're given λ instead of f. A photon also carries momentum (despite having no mass):
p=cE=λh
This follows from special relativity: E2=(pc)2+(mc2)2, and for a photon m=0, so E=pc, hence p=h/λ.
- h=6.626×10−34 J⋅s (Planck's constant)
- c=3.0×108 m/s (speed of light)
- For atomic-scale problems use electronvolts: 1 eV=1.602×10−19 J
A common exam trap
E=hf means higher frequency = more energy per photon — true. But a brighter light does not mean higher photon energy. Intensity is the number of photons per second per area; each photon still carries hf.
Do not confuse intensity (number of photons) with frequency (energy per photon). They are independent.
Final answer: E=hf
Concept: Photon Energy — The energy of an incident photon is E=hν. Part of it goes into overcoming the work function ϕ, and the remainder appears as the maximum kinetic energy Kmax of the photoelectron.
Step 1: Photon energy
E=hν=(6.63×10−34)(6×1014)=3.978×10−19 J
Convert to eV: E=1.6×10−193.978×10−19=2.486 eV
Step 2: Maximum kinetic energy
Kmax=E−ϕ=2.486−2.14=0.346 eV
In joules: Kmax=0.346×1.6×10−19=5.536×10−20 J
Step 3: Stopping potential
V0=eKmax=0.346 V
Step 4: Maximum speed
Kmax=21mvmax2⟹vmax=m2Kmax
vmax=9.1×10−312×5.536×10−20=3.49×105 m/s
- Kmax=0.346 eV,
- V0=0.346 V,
- vmax=3.49×105 m/s
Using Einstein’s photoelectric equation, the maximum kinetic energy is found from the difference between the incident photon energy and the work function. The stopping potential is that kinetic energy divided by the electron charge, and the maximum speed comes from the kinetic energy formula. The answers are: (a) 0.345 eV,
(b) 0.345 V,
(c) 3.48×105 m/s.
The Core Idea: Photon Energy and the Photoelectric Effect
When light hits a metal surface, it behaves as a stream of particles — photons. Each photon carries a quantum of energy given by E=hf, where h is Planck’s constant and f is the frequency. For an electron to be ejected, the photon must supply enough energy to overcome the work function ϕ — the minimum energy needed to free an electron from the metal surface.
Any extra energy beyond ϕ appears as the maximum kinetic energy of the emitted electron. This is Einstein’s photoelectric equation:
Kmax=hf−ϕ
The stopping potential V0 is the voltage that just stops the most energetic electrons — it’s directly related to Kmax by eV0=Kmax. And once we know Kmax in joules, the maximum speed follows from Kmax=21mvmax2.
Let’s apply this step by step.
Step 1: Find the photon energy
The incident light has frequency f=6×1014 Hz. Planck’s constant is h=6.63×10−34 J⋅s.
Photon energy in joules:
E=hf=(6.63×10−34)(6×1014)=3.978×10−19 J
We’ll need this in electronvolts too. Since 1 eV=1.6×10−19 J:
E=1.6×10−193.978×10−19=2.486 eV
A quick check: the product hf in eV can be found using h=4.14×10−15 eV⋅s. Then E=(4.14×10−15)(6×1014)=2.484 eV — essentially the same.
Step 2: Maximum kinetic energy (part a)
Work function ϕ=2.14 eV. Using Einstein’s equation:
Kmax=hf−ϕ=2.486 eV−2.14 eV=0.346 eV
Rounding to three significant figures (matching the given data):
Kmax=0.345 eV
A common mistake is to forget that hf and ϕ must be in the same units. Here both are in eV, so subtraction is straightforward. If you work in joules, convert ϕ first: ϕ=2.14×1.6×10−19=3.424×10−19 J, then Kmax=(3.978−3.424)×10−19=0.554×10−19 J, which equals 0.346 eV — same result.
Step 3: Stopping potential (part b)
The stopping potential V0 satisfies eV0=Kmax. Since Kmax is in eV, the numerical value of V0 in volts is the same:
V0=eKmax=0.345 V
Step 4: Maximum speed (part c)
First convert Kmax to joules:
Kmax=0.345 eV×1.6×10−19 J/eV=5.52×10−20 J
Electron mass m=9.1×10−31 kg. From Kmax=21mvmax2:
vmax=m2Kmax=9.1×10−312×5.52×10−20
Calculate inside the square root:
9.1×10−311.104×10−19=1.213×1011
Taking square root:
vmax=1.213×1011=3.48×105 m/s
This speed is about 0.1% of the speed of light — non-relativistic, so the classical kinetic energy formula is perfectly valid.
(a) Maximum kinetic energy is 0.345 eV, (b) stopping potential is 0.345 V, and (c) maximum speed is 3.48×105 m/s.
Method: Einstein's Photoelectric Equation
This problem is solved using Einstein's photoelectric equation, which states that the incident photon energy is used partly to overcome the work function and the remainder appears as the maximum kinetic energy of the emitted electron.
Step 1: Write down the given data
| Quantity | Value |
|---|---|
| Work function ϕ | 2.14 eV |
| Frequency of incident light f | 6×1014 Hz |
| Planck's constant h | 6.63×10−34 J⋅s |
| 1 eV | 1.6×10−19 J |
| Mass of electron me | 9.1×10−31 kg |
Step 2: Calculate the incident photon energy
Photon energy E=hf
E=(6.63×10−34)(6×1014)=3.978×10−19 J
Convert to eV:
E=1.6×10−193.978×10−19=2.486 eV
Always check whether the photon energy exceeds the work function — only then will photoemission occur. Here 2.486 eV>2.14 eV, so emission is possible.
Step 3: Find maximum kinetic energy (part a)
Einstein's photoelectric equation:
Kmax=hf−ϕ
Kmax=2.486−2.14=0.346 eV
In joules:
Kmax=0.346×1.6×10−19=5.536×10−20 J
Kmax=hf−ϕ
Step 4: Find stopping potential (part b)
The stopping potential V0 is related to Kmax by:
Kmax=eV0
V0=eKmax=e0.346 eV=0.346 V
When Kmax is in eV, the stopping potential in volts is numerically equal to Kmax in eV. So V0=0.346 V directly.
Step 5: Find maximum speed (part c)
Use kinetic energy in joules:
Kmax=21mevmax2
vmax=me2Kmax
vmax=9.1×10−312×5.536×10−20
vmax=1.216×1011=3.487×105 m/s
Final Answers
(a) Maximum kinetic energy: 0.346 eV (or 5.54×10−20 J)
(b) Stopping potential: 0.346 V
(c) Maximum speed: 3.49×105 m/s
Common Mistakes Students Make on This Photon Energy Problem
Mistake 1: Forgetting to convert units before using formulas
The work function is given in eV, but the Planck constant h is usually taken in J·s (6.63×10−34 J⋅s). Students often plug 2.14 eV directly into Kmax=hf−ϕ without converting everything to joules first.
How to avoid: Always check unit consistency. Convert the work function from eV to joules using 1 eV=1.6×10−19 J:
ϕ=2.14×1.6×10−19=3.424×10−19 J
Now compute hf in joules:
hf=(6.63×10−34)(6×1014)=3.978×10−19 J
Then Kmax=3.978×10−19−3.424×10−19=5.54×10−20 J.
If you keep ϕ in eV and hf in joules, you'll get a meaningless number. Always work in a single unit system — joules is safest for kinetic energy and speed calculations.
Mistake 2: Confusing stopping potential with maximum kinetic energy
Students sometimes write V0=Kmax directly, forgetting that stopping potential is related by eV0=Kmax.
How to avoid: Remember the definition: stopping potential is the voltage that just stops the most energetic electrons. The work done by the electric field (eV0) equals the maximum kinetic energy lost. So:
V0=eKmax
Using Kmax=5.54×10−20 J:
V0=1.6×10−195.54×10−20=0.346 V
If you already have Kmax in eV, then V0 in volts is numerically equal to Kmax in eV. Here Kmax=0.346 eV, so V0=0.346 V — a handy shortcut.
Mistake 3: Using the wrong mass for the electron in the speed calculation
Students sometimes use the mass of a proton or forget to square the speed properly in K=21mv2.
How to avoid: The electron mass is me=9.1×10−31 kg. Rearranging:
vmax=me2Kmax=9.1×10−312×5.54×10−20
Compute step by step:
- 2Kmax=1.108×10−19
- Divide by me: 9.1×10−311.108×10−19=1.218×1011
- Take square root: vmax=3.49×105 m/s
| Quantity | Value | Unit |
|----------|-------|------|
| Kmax | 5.54×10−20 | J |
| V0 | 0.346 | V |
| vmax | 3.49×105 | m/s |
Mistake 4: Forgetting that photoemission requires hf≥ϕ
Some students attempt the problem even when the photon energy is below the work function, getting a negative kinetic energy.
How to avoid: First check if emission is possible. Here hf=3.978×10−19 J and ϕ=3.424×10−19 J, so hf>ϕ — emission occurs. If hf<ϕ, simply state "no photoemission" and stop.
Mistake 5: Rounding intermediate values too aggressively
Rounding Kmax to 5.5×10−20 J early can throw off the speed calculation significantly because of the square root.
How to avoid: Keep at least 3 significant figures throughout, and round only the final answer. Use h=6.63×10−34 consistently (not 6.6×10−34).
Final answers:
- (a) Kmax=5.54×10−20 J (or 0.346 eV)
- (b) V0=0.346 V
- (c) vmax=3.49×105 m/s
Showing the 12 most recent of 30 on this concept.
- CBSE 2026Set ANNUAL1 markQ.If the wavelength of a photon is halved, then its frequency will become ______.
›Reveal solutionSolution
Since nu = c/lambda for a photon, halving lambda directly doubles nu (c is a universal constant).
A photon's frequency and wavelength are related by nu = c/lambda, where c (speed of light) is fixed. If lambda is halved (lambda -> lambda/2), then nu = c/(lambda/2) = 2(c/lambda), i.e. the frequency becomes twice its original value.
✓Final answerdoubled (2x the original frequency).
- CBSE 2026Set ANNUAL1 markMCQQ.The mass of a photon is:(a) h/v(b) hc/λ(c) h/λ(d) hν/c²
›Reveal solutionSolution
A photon's energy is E=hν; equating this to E=mc2 gives its (relativistic/effective) mass m=hν/c2.
A photon has zero rest mass but carries energy E=hν (Planck's relation) and momentum p=h/λ=E/c. Using Einstein's mass-energy equivalence E=mc2 for the energy it carries while in motion, its effective mass is m=c2E=c2hν. The other options are quantities in disguise: hc/λ=hν is the photon's energy, not its mass, and h/λ is its momentum (p=h/λ), not its mass.
✓Final answer(d) hν/c2
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: The rest mass of photon is ______.
›Reveal solutionSolution
A photon's rest mass is zero.
A photon is a quantum of electromagnetic radiation that always travels at the speed of light c in vacuum. According to relativity, any particle moving at speed c must have zero rest mass; otherwise its energy would be infinite. A photon does have energy (E = hν) and momentum (p = hν/c), but its rest mass (mass measured when at rest) is zero.
✓Final answerzero.
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: The value of Planck's constant is ______.
›Reveal solutionSolution
Planck's constant h ≈ 6.63 × 10⁻³⁴ J·s.
Planck's constant h relates the energy of a photon to its frequency by E = hν. Its accepted value is
h = 6.63 × 10⁻³⁴ joule-second (J·s).
It is one of the fundamental constants of nature and appears throughout quantum physics.
✓Final answer6.63 × 10⁻³⁴ J·s.
- CBSE 2025Set 55/4/11 markMCQQ.A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true? (A) The blue beam has more number of photons than the red beam. (B) The red beam has more number of photons than the blue beam. (C) Wavelength of red light is lesser than the wavelength of blue light. (D) The blue light beam has lesser energy per photon than that in the red light beam.
›Reveal solutionSolution
Since blue photons carry more energy than red photons, equal-intensity beams require more red photons to match the same total power. The red beam has more photons.
The key to this problem lies in understanding what "intensity" means and how photon energy depends on wavelength.
Intensity measures the power (energy per unit time) delivered per unit area. When two beams have equal intensities, they carry the same total energy per second through the same cross-sectional area, regardless of color.
Each photon carries energy E=hν=λhc, where h is Planck's constant, c is the speed of light, and λ is the wavelength. Blue light has a shorter wavelength than red light (λblue<λred), which means blue photons are individually more energetic than red photons.
If the total power delivered by both beams is the same, but blue photons pack more energy each, then fewer blue photons are needed to deliver that power. Conversely, more red photons are required to compensate for their lower individual energy.
Let me work through this quantitatively:
- Express intensity in terms of photon count. If n photons pass through area A in time t, the intensity is:
I=A⋅tTotal energy=A⋅tn⋅Ephoton=A⋅tn⋅hc/λ
- Set up the equal-intensity condition. For red and blue beams with equal intensities:
Ired=Iblue
A⋅tnred⋅hc/λred=A⋅tnblue⋅hc/λblue
- Simplify to find the photon ratio:
nred⋅λred1=nblue⋅λblue1
nbluenred=λblueλred
- Apply the wavelength relationship. Since red light has a longer wavelength than blue light (λred>λblue):
nbluenred>1⟹nred>nblue
Now let's check each option:
- (A) Claims blue has more photons — false, we just showed the opposite.
- (B) Claims red has more photons — true, matches our derivation.
- (C) Claims red wavelength is less than blue — false, red has longer wavelength.
- (D) Claims blue photons have less energy — false, Eblue=hc/λblue>hc/λred=Ered.
TipA quick mnemonic: "Lower energy photons need higher numbers" — to match the same total power, the beam with less energetic photons must have more of them.
✓Final answerThe correct option is (B): the red beam has more photons than the blue beam.
- CBSE 2025Set 55/5/11 markMCQQ.Which of the following electromagnetic waves has photons of the largest momentum? (A) X-rays (B) AM radio waves (C) Microwaves (D) TV waves
›Reveal solutionSolution
Photon momentum is p=λh, so the wave with the shortest wavelength has the largest momentum. Among the options, X-rays have the shortest wavelength, hence the largest photon momentum.
Concept & Intuition
The momentum of a photon is not like the momentum of a massive particle. For a photon, momentum is purely a wave property, given by the de Broglie relation:
p=λh
where h is Planck’s constant and λ is the wavelength. This means: shorter wavelength → larger momentum. There is no dependence on amplitude or intensity — only wavelength matters.
So the question reduces to: which of these electromagnetic waves has the shortest wavelength? Let’s recall the electromagnetic spectrum order from longest to shortest wavelength:
- Radio waves (including AM and TV) — longest wavelengths (metres to kilometres)
- Microwaves — centimetres to millimetres
- Infrared — micrometres
- Visible light — hundreds of nanometres
- Ultraviolet — tens of nanometres
- X-rays — picometres to nanometres
- Gamma rays — sub-picometre
Watch outA common mistake is to think that higher frequency means higher energy (true), but then incorrectly assume that momentum depends on something else like the wave’s “penetrating power” or “ionising ability”. Stick to p=h/λ — it’s the only formula that matters here.
Step-by-step solution
-
Write the momentum formula
For any photon, p=λh. Since h is constant, p∝λ1.
-
Identify the wavelengths of each option
- AM radio waves: wavelength ≈100 m to 1000 m (longest)
- TV waves: wavelength ≈0.1 m to 10 m (still radio band)
- Microwaves: wavelength ≈1 mm to 30 cm
- X-rays: wavelength ≈0.01 nm to 10 nm (shortest among these)
-
Compare
Since p∝1/λ, the smallest λ gives the largest p. X-rays have the smallest wavelength by many orders of magnitude.
-
Conclude
X-ray photons carry the largest momentum.
TipYou don’t need to memorise exact numbers — just remember the order of the EM spectrum from longest to shortest wavelength: Radio → Microwave → Infrared → Visible → UV → X-ray → Gamma. The one furthest to the right among the options wins.
✓Final answerThe correct option is (A) X-rays.
- CBSE 2025Set D1 markMCQQ.What is the energy of a photon with a wavelength of 500 nm? ( Use c = 3 × 10^8 m/s and h = 6.626 × 10^-34 Js ) (A) 4 × 10^-19 J (B) 2.5 × 10^-19 J (C) 1.2 × 10^-18 J (D) 6.6 × 10^-19 J
›Reveal solutionSolution
Photon energy E = hc/λ ≈ 4 × 10⁻¹⁹ J for λ = 500 nm.
The energy of a photon is
E=λhc
Substitute h = 6.626×10⁻³⁴ J·s, c = 3×10⁸ m/s, λ = 500 nm = 500×10⁻⁹ m:
E=500×10−9(6.626×10−34)(3×108)
E=5×10−71.9878×10−25=3.98×10−19 J
This rounds to 4 × 10⁻¹⁹ J.
✓Final answer(A) 4 × 10⁻¹⁹ J.
- CBSE 2025Set A1 markQ.Match Column 'A' item 'Frequency of light' with the correct option from Column 'B' and write the correct pair. Column 'B' options:(i) Minimum energy to emit electrons from the surface(ii) Minimum frequency to emit electrons from the surface(iii) Frequency of photon(iv) Number of photons(v) Moving particle(vi) Photon(vii) Einstein.
›Reveal solutionSolution
Frequency of light corresponds to option (iii): frequency of photon.
In the photon (particle) picture of light proposed by Einstein, a beam of light of frequency ν is regarded as a stream of photons, each carrying energy E = hν — so the 'frequency of light' (a wave concept) and the 'frequency of the photon' (used to compute each photon's quantum of energy) are simply the same physical quantity viewed from the two complementary (wave/particle) descriptions of light. Hence 'Frequency of light' matches '(iii) Frequency of photon'.
✓Final answerFrequency of light → (iii) Frequency of photon.
- CBSE 2025Set ANNUAL1 markQ.Electron volt (eV) is the unit of ................. (fill in the blank)
›Reveal solutionSolution
The electron volt is a convenient small unit of energy, widely used in atomic and nuclear physics.
One electron volt is defined as the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt:
1 eV=1.6×10−19 J
Because atomic, photon, and nuclear energies are typically tiny fractions of a joule, the eV (and its multiples keV, MeV) is the standard convenient energy unit in this domain.
✓Final answerEnergy.
- CBSE 2025Set ANNUAL1 markQ.A blue lamp mainly emits light of wavelength 4500A∘. The lamp is rated at 150 W and 8% of energy is emitted as visible light. How many photons are emitted by lamp per second?
›Reveal solutionSolution
Visible-light power = 8% of 150 W; divide by the energy of one photon at 4500 Å.
Power emitted as visible light =8% of 150W =0.08×150=12W.
Energy of one photon at λ=4500A˚=4.5×10−7m:
E=λhc=4.5×10−76.63×10−34×3×108≈4.42×10−19 J
Number of photons emitted per second:
n=EP=4.42×10−1912≈2.71×1019 photons/s
✓Final answerAbout 2.71×1019 photons are emitted per second.
- CBSE 2025Set ANNUAL1 markMCQQ.The momentum of a photon of energy h.nu is(i) h.nu(ii) h.nu/c(iii) h.nu.c(iv) h/nu
›Reveal solutionSolution
Photon momentum p = E/c = h(nu)/c.
A photon of frequency ν carries energy E=hν. Being a massless quantum that moves at the speed of light, its momentum is p=E/c. Therefore p=chν (equivalently p=h/λ since c=νλ).
✓Final answer(ii) h.nu/c.
- CBSE 2024Set ANNUAL1 markMCQQ.The momentum (p) of photon is -(a) h/λ(b) λ/h(c) hC/λ(d) hλ
›Reveal solutionSolution
A photon's momentum follows from combining its energy E = hc/λ with the relativistic relation E = pc for a massless particle.
A photon of frequency ν has energy E=hν=λhc (since c=νλ).
A photon is massless and travels at speed c, so by the relativistic energy-momentum relation for a massless particle, E=pc. Equating the two expressions for E:
pc=λhc⟹p=λh
✓Final answer(a) h/λ.
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