Q.Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
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🔒 Start your 14-day free trial to unlock the full solution →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 …
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): …
Concept: Photon Energy — each photon carries energy E=hν=λhc and momentum p=λh.
(a)
Wavelength λ=632.8 nm=6.328×10−7 m.
Energy:
E=λhc=6.328×10−7(6.626×10−34)(3×108)=3.14×10−19 J
Momentum:
p=λh=6.328×10−76.626×10−34=1.05×10−27 kg m/s
(b)
Power P=9.42 mW=9.42×10−3 J/s.
Number of photons per second:
n=EP=3.14×10−199.42×10−3=3.00×1016 s−1
(c) …
The problem connects photon energy, momentum, and number flux to macroscopic laser power. Each photon carries energy E=hc/λ and momentum p=h/λ; the number of photons per second is power divided by photon energy; and the hydrogen atom’s speed for equal momentum comes from p=mHv.
Concept and Intuition
A laser beam is a stream of photons. Even though the beam looks continuous, its power is the sum of the energies of individual photons arriving per second. Each photon, being a quantum of light, has energy proportional to its frequency and momentum inversely proportional to its wavelength — a direct consequence of de Broglie’s relation and Planck’s law.
The key is to treat the macroscopic power (9.42 mW) as the product of the number of photons per second and the energy per photon. For part (c), we simply equate the photon’s momentum to the classical momentum of a hydrogen atom and solve for its speed.
Step-by-step solution
1. Photon energy from wavelength
The energy of a single photon is given by the Planck-Einstein relation:
E=hf=λhc
where
h=6.626×10−34 J⋅s (Planck’s constant),
c=3.00×108 m/s,
λ=632.8 nm=632.8×10−9 m.
Substitute:
E=632.8×10−9(6.626×10−34)(3.00×108)
First compute numerator: 6.626×3.00=19.878, so 19.878×10−26 J⋅m.
Divide by 632.8×10−9:
E=632.8×10−919.878×10−26=632.819.878×10−17
632.819.878≈0.03141, so
E≈3.141×10−19 J
A quick check: visible photons have energies around 10−19 J, so this result is reasonable.
2. Photon momentum
For a photon, momentum is:
p=λh
Substitute:
p=632.8×10−96.626×10−34=632.86.626×10−25
632.86.626≈0.01047, so
p≈1.047×10−27 kg⋅m/s
Do not use p=E/c here unless you keep units consistent — it gives the same result but is one extra step. The direct h/λ is simpler.
3. Number of photons per second
Power P=9.42 mW=9.42×10−3 J/s.
If each photon carries energy E, then the number of photons arriving per second is:
n=EP
Substitute:
n=3.141×10−199.42×10−3 …
Method: Photon Energy–Momentum Relations and Power–Photon Rate Conversion
This problem uses two core ideas: the Planck–Einstein relations for a single photon, and the fact that the total power of a beam equals the energy per photon times the number of photons per second.
(a) Energy and momentum of each photon
Step 1 – Convert wavelength to metres.
632.8 nm=632.8×10−9 m=6.328×10−7 m.
Step 2 – Energy of one photon.
Use E=λhc, where h=6.626×10−34 J⋅s and c=3.00×108 m/s.
E=6.328×10−7(6.626×10−34)(3.00×108)
First compute numerator: 6.626×3.00=19.878, with powers 10−34+8=10−26, so 1.9878×10−25 J⋅m.
Divide: 6.328×10−71.9878×10−25=3.141×10−19 J.
You can also express this in electronvolts: 1 eV=1.602×10−19 J, so E≈1.96 eV — a typical value for red laser light.
Step 3 – Momentum of one photon.
Use p=λh.
p=6.328×10−76.626×10−34=1.047×10−27 kg⋅m/s
E=λhc,p=λh
(b) Number of photons per second arriving at the target
Step 1 – Power is energy per second.
P=9.42 mW=9.42×10−3 J/s.
Step 2 – Let n be the number of photons per second.
Then P=n×E (energy of one photon).
n=EP=3.141×10−199.42×10−3
Step 3 – Divide.
9.42/3.141≈3.00, and 10−3/10−19=1016.
n=3.00×1016 photons/s …
Here are the most common mistakes students make on this exact problem, and how to avoid each one.
1. Forgetting to convert units (nm → m, mW → W)
The most frequent error. Wavelength is given in nanometres (1 nm=10−9 m) and power in milliwatts (1 mW=10−3 W). Plugging 632.8 directly into E=hc/λ gives a completely wrong energy — off by nine orders of magnitude.
How to avoid: Before writing any formula, convert every quantity to SI base units. Write it explicitly:
λ=632.8×10−9 m=6.328×10−7 m
P=9.42×10−3 W
Make this the first line of your solution.
2. Using the wrong formula for photon momentum
Many students write p=mc or p=E/c2 — both are incorrect for a photon. A photon has zero rest mass, so p=mv does not apply.
How to avoid: Memorise the two equivalent forms for photon momentum:
p=λhorp=cE
Use the first one here since you already have λ. It’s direct and avoids any confusion with mass.
3. Confusing energy per photon with total power
Part (b) asks for the number of photons per second. A common mistake is to divide power by c or by λ, or to use E=hf but forget to find f from λ.
How to avoid: Remember the chain:
Number of photons per second=Energy of one photonTotal energy per second (power)
So first compute Ephoton=hc/λ, then:
n=EphotonP
Do not skip the intermediate step — write Ephoton explicitly before dividing.
4. Using the wrong mass for the hydrogen atom in part (c)
Some students use the mass of a proton (1.67×10−27 kg) or the mass of an electron. The problem says hydrogen atom, so you need the mass of the entire atom — approximately the mass of a proton plus an electron, which is essentially 1.67×10−27 kg (the electron contributes negligibly, but using the proton mass is acceptable here). A bigger mistake is using the mass of a hydrogen molecule (H2) or forgetting mass entirely and trying v=p/c.
How to avoid: Write down the known mass of a hydrogen atom: …
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) …
- 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 q …
- 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 …
- 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).
…
- 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 …
- 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) …
- 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) …
- 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 t …
- 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 …
- 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
…
- 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.
…
- 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).
…
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