Skip to content
Problems · Problem 2.7

Q.A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

Andhra Pradesh BieapTextbookSubjective· 2mImportance★★★★★est
5% · 7/140 Questions
✓ Free question

Each photon carries energy E=hcλE = \frac{hc}{\lambda}; dividing the bulb's power by the energy per photon gives the photon emission rate of 2.01×10202.01 \times 10^{20} photons per second.

Why this approach works

Power tells us energy per unit time. A 100 W bulb delivers 100 joules every second. If the bulb emits monochromatic light, every photon carries the same energy determined by its wavelength. The number of photons emitted per second is simply the total energy per second divided by the energy carried by one photon.

The energy of a single photon depends on its frequency (or equivalently, its wavelength) through Planck's relation:

Ephoton=hν=hcλE_{\text{photon}} = h\nu = \frac{hc}{\lambda}

where hh is Planck's constant, cc is the speed of light, and λ\lambda is the wavelength.

Step-by-step calculation

1. Identify the given quantities

The bulb's power is P=100 W=100 J/sP = 100 \text{ W} = 100 \text{ J/s}, meaning it emits 100 joules of energy every second. The wavelength of the emitted light is λ=400 nm=400×10−9 m\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}.

2. Calculate the energy of one photon

Using the photon energy formula:

Ephoton=hcλE_{\text{photon}} = \frac{hc}{\lambda}

Substitute the constants:

  • Planck's constant: h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34} \text{ J·s}
  • Speed of light: c=3×108 m/sc = 3 \times 10^8 \text{ m/s}
  • Wavelength: λ=400×10−9 m\lambda = 400 \times 10^{-9} \text{ m}

Ephoton=(6.626×10−34)(3×108)400×10−9E_{\text{photon}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}}

Ephoton=19.878×10−26400×10−9=19.878×10−264×10−7E_{\text{photon}} = \frac{19.878 \times 10^{-26}}{400 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}}

Ephoton=4.9695×10−19 JE_{\text{photon}} = 4.9695 \times 10^{-19} \text{ J}

Rounding appropriately, Ephoton≈4.97×10−19 JE_{\text{photon}} \approx 4.97 \times 10^{-19} \text{ J}.

3. Find the number of photons emitted per second

The number of photons NN emitted per second is the total energy emitted per second divided by the energy per photon:

N=PEphoton=1004.97×10−19N = \frac{P}{E_{\text{photon}}} = \frac{100}{4.97 \times 10^{-19}}

N=2.012×1020 photons/sN = 2.012 \times 10^{20} \text{ photons/s}

Tip

For quick estimates, remember that visible light photons (400–700 nm) carry energies on the order of 10−1910^{-19} J. A 1 W source emits roughly 101810^{18} to 101910^{19} photons per second.

Watch out

This calculation assumes all the bulb's electrical power is converted to light. Real incandescent bulbs are inefficient (most energy becomes heat), so the actual number of photons would be much smaller. The problem implicitly asks for the idealized case where 100 W of optical power is emitted.

✓Final answer

The bulb emits 2.01×1020\boxed{2.01 \times 10^{20}} photons per second.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question · 2.3.2Problem 2.6 — Calculate energy of one mole of photons of radiation whose frequency i…→Next question · 2.3.2Problem 2.8 — When electromagnetic radiation of wavelength 300 nm falls on the surfa…