Q.How many photons per second emanate from a 50 mW laser of wavelength 640 nm?
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
Ephoton=hc/λ≈3.106×10−19 J. n=P/Ephoton=0.05/3.106×10−19≈1.61×1017 s−1.
n≈1.61×1017 photons per second
Step 1. Convert the laser's power: P=50 mW=0.05 W.
Step 2. The energy of one photon at λ=640 nm is E=λhc=640×10−9(6.626×10−34)(3×108)≈3.106×10−19 J.
Step 3. The number of photons emitted per second is n=EP=3.106×10−190.05≈1.61×1017 s−1.
n≈1.61×1017 photons per second
Divide the laser's power (converted to watts) by the energy of a single photon at the given wavelength.
- Forgetting to convert 50 mW to 0.05 W before dividing.
- Using the wavelength in nanometres directly without converting to metres.
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.