A photon has energy given by E = hf, where h = 6.63 × 10⁻³⁴ J·s and f = 5 × 10¹⁴ Hz. What is the energy of the photon in joules? - Nelissen Grade advocaten

Understanding the Context

where:

• E = energy of the photon (in joules, J),
  
• h = Planck’s constant = 6.63 × 10⁻³⁴ J·s,
  
• f = frequency of the electromagnetic wave.

What Does This Equation Mean?

This simple yet powerful formula tells us that a photon’s energy is directly proportional to its frequency. Higher-frequency waves—like blue light or gamma rays—carry more energy than lower-frequency waves such as radio waves or infrared radiation.

Converting Frequency to Energy

Key Insights

Let’s apply the formula using real values.
Given:

• Planck’s constant, h = 6.63 × 10⁻³⁴ J·s
  
• Frequency, f = 5 × 10¹⁴ Hz

Now plug into E = hf:
E = (6.63 × 10⁻³⁴ J·s) × (5 × 10¹⁴ Hz)

Multiplying the numbers:
E = 6.63 × 5 × 10⁻³⁴⁺¹⁴
E = 33.15 × 10⁻²⁰ J
E = 3.315 × 10⁻¹⁹ J

Summary

The energy of a photon with frequency 5 × 10¹⁴ Hz is approximately 3.315 × 10⁻¹⁹ joules. This precise calculation highlights how quantum mechanics quantifies light not as continuous waves, but as discrete packets of energy—photons—each carrying a measurable amount tied directly to their frequency.

Final Thoughts

Whether studying optical physics, quantum mechanics, or photon-based technologies like lasers or solar cells, understanding E = hf is essential. It bridges the abstract concept of wave-particle duality with concrete, measurable energy values.