Q.Monochromatic light of frequency 6.0×1014 Hz is produced by a laser. The power emitted is 2.0×10−3 W.
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 — each photon carries a discrete quantum of energy E=hν, where h is Planck’s constant and ν is the frequency.
(a) Energy of one photon:
E=hν=(6.63×10−34)(6.0×1014)=3.978×10−19 J
(b) Power P is energy per second. If n photons are emitted each second, total energy per second is nE=P. Hence:
n=EP=3.978×10−192.0×10−3≈5.03×1015
- The energy of a photon is 3.98×10−19 J;
- the number of photons emitted per second is 5.0×1015.
The energy of a single photon is found using E=hν, giving 3.98×10−19 J. The number of photons emitted per second is the total power divided by the photon energy, yielding 5.0×1015 photons/s.
Why Photon Energy Matters Here
Light is not a continuous stream of energy — it comes in discrete packets called photons. Each photon carries a specific energy that depends only on the frequency (or wavelength) of the light, not on the intensity. The laser's power tells us how much total energy is delivered per second. To find how many photons leave the laser each second, we simply divide the total energy per second (power) by the energy carried by one photon.
This is a clean, two-step problem: first find the energy of one photon, then count how many such photons make up the total power.
Step-by-Step Solution
1. Energy of a single photon
The energy E of one photon is given by the Planck-Einstein relation:
E=hν
where
h=6.626×10−34 J⋅s (Planck's constant)
ν=6.0×1014 Hz (frequency)
Substitute:
E=(6.626×10−34)×(6.0×1014)
E=3.9756×10−19 J
Rounding to two significant figures (matching the given data):
E≈3.98×10−19 J
Ephoton=hν
If you ever forget the value of h, remember it's roughly 6.63×10−34 J⋅s. For quick mental checks: light of frequency 5×1014 Hz (yellow-green) has photon energy about 3.3×10−19 J.
2. Number of photons emitted per second
Power P is energy per unit time. If each photon carries energy E, then the number of photons emitted per second n satisfies:
P=n×E
So:
n=EP
Given P=2.0×10−3 W (which is 2.0×10−3 J/s):
n=3.9756×10−192.0×10−3
n=5.03×1015 photons/s
Rounding to two significant figures:
n≈5.0×1015 photons/s
A common mistake is to forget that power is already in joules per second — no extra conversion is needed. Also, be careful with exponents: 10−3 divided by 10−19 gives 1016, not 10−22.
The energy of a photon is 3.98×10−19 J and the number of photons emitted per second is 5.0×1015 photons/s.
Method: Photon Energy Approach (using Planck's relation)
This problem uses the fundamental idea that light energy comes in discrete packets called photons. The energy of each photon depends only on the frequency of the light, not on the power. Power tells us how much total energy is delivered per second, so dividing that by the energy per photon gives the number of photons per second.
(a) Energy of a single photon
Step 1: Recall Planck's relation — the energy of one photon is directly proportional to its frequency:
E=hf
where h=6.63×10−34 J⋅s (Planck's constant) and f is the frequency in hertz.
Step 2: Substitute the given frequency:
E=(6.63×10−34)(6.0×1014)
Step 3: Multiply the numbers and the powers of ten separately:
6.63×6.0=39.78
10−34×1014=10−20
So E=39.78×10−20 J=3.978×10−19 J
Always check the exponent: 10−34×1014=10−20, not 10−48 — a common slip.
Step 4: Round to two significant figures (since the given frequency has two significant figures):
E=4.0×10−19 J
(b) Number of photons emitted per second
Step 1: Understand what power means. Power P=2.0×10−3 W means the source delivers 2.0×10−3 joules of energy each second.
Step 2: If each photon carries E joules, then the number of photons emitted per second, n, is:
n=energy per photontotal energy per second=EP
Step 3: Substitute the values:
n=4.0×10−192.0×10−3
Step 4: Divide the coefficients and subtract the exponents:
4.02.0=0.5
10−1910−3=1016
So n=0.5×1016=5.0×1015
Dividing powers of ten: 10−3÷10−19=10−3−(−19)=1016. The minus of a minus becomes plus.
n=5.0×1015 photons per second
Final answers:
- Energy of one photon = 4.0×10−19 J
- Number of photons emitted per second = 5.0×1015
Common Mistakes & How to Avoid Them
Mistake 1: Using the wrong formula for photon energy
Students often confuse E=hf with E=λhc. Both are correct, but the first is direct when frequency is given. The second requires an extra step (converting frequency to wavelength) and introduces more places for error — like using the wrong value for c or forgetting to convert units.
How to avoid: When frequency is given, use E=hf directly. Only switch to E=hc/λ when wavelength is provided. Memorise both forms but pick the one that matches the data.
If the problem gives frequency, your first instinct should be E=hf. No conversions needed.
Mistake 2: Forgetting the value of Planck's constant or using the wrong one
Planck's constant h is 6.63×10−34 J⋅s. Some students use 6.6×10−34 (acceptable in some boards, but risky) or accidentally use h=4.14×10−15 eV⋅s when the answer is expected in joules.
How to avoid: Write down h=6.63×10−34 J⋅s at the top of your working. If the question asks for energy in joules (which it usually does unless specified), stick to this value. If you must use the eV version, convert at the end — don't mix units mid-calculation.
Mistake 3: Incorrect exponent handling in part (a)
The calculation is:
E=(6.63×10−34)(6.0×1014)
Students often add exponents incorrectly: 10−34×1014=10−20, not 10−48 or 10−20 with a sign error. Also, they sometimes forget to multiply the coefficients: 6.63×6.0≈39.78, not 3.978.
How to avoid: Separate the calculation into two parts:
- Multiply the coefficients: 6.63×6.0=39.78
- Add the exponents: −34+14=−20
- Combine: 39.78×10−20=3.978×10−19
Then round to appropriate significant figures (here, two significant figures from the given data gives 4.0×10−19 J).
10−34×1014=10−20, not 10−48 (that's multiplying exponents instead of adding them). This is the single most common exponent error.
Mistake 4: Confusing power with energy in part (b)
Power P=2.0×10−3 W means 2.0×10−3 J of energy is emitted per second. Some students treat power as the total energy or forget that it's already a rate.
How to avoid: Write down what power means: P=tEtotal. For t=1 s, Etotal=P×1=P. So the number of photons per second is:
n=energy per photontotal energy per second=EP
Mistake 5: Dividing in the wrong order
Students sometimes compute E/P instead of P/E, getting a tiny fraction instead of a large number.
How to avoid: Check the units. You want photons per second, which has units of s−1. P has units J/s, E has units J. So P/E gives JJ/s=s−1, which is correct. E/P gives seconds — a time, not a rate.
Unit check: JW=JJ/s=s−1. Always verify your formula by checking what units it produces.
Mistake 6: Arithmetic errors in part (b)
n=4.0×10−192.0×10−3=0.5×1016=5.0×1015
Common errors: dividing coefficients as 2.0/4.0=0.5 but then writing 0.5×10−16 (sign error on exponent), or forgetting that 10−3/10−19=1016.
How to avoid: Again, separate coefficient and exponent:
- Coefficients: 2.0/4.0=0.5
- Exponents: −3−(−19)=−3+19=16
- Combine: 0.5×1016=5.0×1015
Final Answers
(a) E=hf=4.0×10−19 J
(b) n=EP=5.0×1015 photons per second
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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