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?
Understanding Photon Energy: Calculating E = hf
In the fascinating world of quantum physics, photons—elementary particles that carry light and electromagnetic radiation—play a crucial role. One fundamental equation defines a photon's energy:
E = hf
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
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.
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.
