Photon Energy and Wavelength Calculator: E = hc/λ Explained

The Planck-Einstein Relation

Light consists of discrete packets of energy called photons. The energy of a single photon is directly related to its frequency and inversely related to its wavelength:

Where:

• E = photon energy (joules or electron volts)
• h = Planck's constant = 6.62607015 × 10⁻³⁴ J·s
• f = frequency (Hz)
• c = speed of light = 299,792,458 m/s
• λ (lambda) = wavelength (meters)

The product hc appears so frequently in calculations that it's useful to know:

• hc = 1.98644568 × 10⁻²⁵ J·m
• hc = 1239.84198 eV·nm (extremely useful for visible light)
• hc = 12398.4198 eV·Å (useful for X-rays)

Understanding the Inverse Relationship

The photon energy formula reveals a crucial inverse relationship between energy and wavelength:

• Shorter wavelength → Higher energy: Gamma rays, X-rays, and ultraviolet light have short wavelengths and high energies
• Longer wavelength → Lower energy: Infrared, microwaves, and radio waves have long wavelengths and low energies

This relationship has profound implications:

• UV light can cause sunburn and skin cancer because its photons have enough energy to damage DNA
• Visible light can trigger chemical reactions in our eyes (vision) and in plants (photosynthesis)
• Infrared light mostly just heats objects because its photons don't have enough energy to break chemical bonds
• X-rays can penetrate tissue because their high-energy photons aren't absorbed by soft matter

Energy Units: Joules vs Electron Volts

Photon energies can be expressed in joules (SI unit) or electron volts (more practical for atomic-scale energies).

The Joule (J)

The SI unit of energy. One joule equals one kilogram meter squared per second squared (1 J = 1 kg·m²/s²). For individual photons, energies are extremely small in joules.

The Electron Volt (eV)

The energy gained by an electron accelerating through a potential difference of 1 volt:

• 1 eV = 1.602176634 × 10⁻¹⁹ J
• 1 keV = 1,000 eV = 1.602 × 10⁻¹⁶ J
• 1 MeV = 1,000,000 eV = 1.602 × 10⁻¹³ J
• 1 GeV = 1,000,000,000 eV = 1.602 × 10⁻¹⁰ J

Electron volts are convenient because visible light photons have energies of about 1.5-3 eV, a much easier number to work with than 10⁻¹⁹ joules.

Converting Between Units

To convert energy from joules to eV:

To convert from eV to joules:

Simplified Formulas for Practical Calculations

Using the value hc = 1239.84 eV·nm, the energy formula becomes remarkably simple:

For wavelength in angstroms (useful for X-rays):

For wavelength in micrometers (useful for infrared):

To find wavelength from energy:

Step-by-Step Calculation Process

To Calculate Photon Energy from Wavelength:

1. Express wavelength in meters (or use the simplified formula with nm)
2. Apply the formula: E = hc/λ = (6.626 × 10⁻³⁴)(3 × 10⁸)/λ
3. Convert to desired units (eV, keV, etc.)

To Calculate Wavelength from Photon Energy:

1. Express energy in joules (or use eV with the simplified formula)
2. Apply the formula: λ = hc/E = (6.626 × 10⁻³⁴)(3 × 10⁸)/E
3. Convert to desired units (nm, Å, etc.)

Worked Examples

Example 1: Green Light (550 nm)

Problem: Calculate the energy of a green light photon with wavelength 550 nm.

Solution using simplified formula:

E = 1239.84 / 550 = 2.254 eV

Verification using full formula:

1. Convert wavelength: 550 nm = 550 × 10⁻⁹ m = 5.5 × 10⁻⁷ m
2. E = hc/λ = (6.626 × 10⁻³⁴)(2.998 × 10⁸)/(5.5 × 10⁻⁷)
3. E = 3.61 × 10⁻¹⁹ J
4. Convert to eV: 3.61 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ = 2.254 eV ✓

Answer: 2.25 eV

Example 2: Red Light (700 nm)

Problem: What is the energy of a red photon at 700 nm?

Solution:

E = 1239.84 / 700 = 1.771 eV

Answer: 1.77 eV

Example 3: Violet Light (400 nm)

Problem: Calculate the energy of violet light at 400 nm.

Solution:

E = 1239.84 / 400 = 3.100 eV

Answer: 3.10 eV

Notice that violet photons have nearly twice the energy of red photons!

Example 4: UV-B Radiation (300 nm)

Problem: What is the energy of UV-B radiation responsible for sunburn?

Solution:

E = 1239.84 / 300 = 4.133 eV

Answer: 4.13 eV

This energy exceeds the bond strength of many organic molecules (typically 2-5 eV), which is why UV-B can damage DNA and cause skin cancer.

Example 5: Medical X-ray (0.05 nm)

Problem: Calculate the energy of a medical X-ray with wavelength 0.05 nm (0.5 Å).

Solution:

E = 1239.84 / 0.05 = 24,797 eV = 24.8 keV

Answer: 24.8 keV

This is a typical energy for diagnostic X-rays. Therapeutic X-rays for cancer treatment use even higher energies (MeV range).

Example 6: Infrared Heat Lamp (10 μm)

Problem: What is the photon energy of infrared radiation at 10 μm?

Solution:

10 μm = 10,000 nm

E = 1239.84 / 10,000 = 0.124 eV = 124 meV

Answer: 0.124 eV (124 meV)

Example 7: Finding Wavelength from Energy

Problem: A photon has energy 6.2 eV. What is its wavelength?

Solution:

λ = 1239.84 / 6.2 = 200 nm

Answer: 200 nm (deep UV, germicidal wavelength)

Example 8: Gamma Ray (1 MeV)

Problem: What wavelength corresponds to a 1 MeV gamma ray?

Solution:

1 MeV = 1,000,000 eV

λ = 1239.84 / 1,000,000 = 0.00124 nm = 1.24 pm

Answer: 1.24 pm (about 1/1000 the size of an atom)

Photon Energy Across the Electromagnetic Spectrum

This table shows typical wavelengths and corresponding photon energies for different parts of the electromagnetic spectrum:

Physical Significance of Photon Energy

Threshold Energies in Physics and Chemistry

Photon energy determines what physical and chemical processes are possible:

The Photoelectric Effect

Einstein's explanation of the photoelectric effect earned him the Nobel Prize. When light shines on a metal surface, electrons are ejected only if the photon energy exceeds the metal's work function (the minimum energy needed to remove an electron):

Typical work functions range from 2 to 6 eV. If photon energy is below the work function, no electrons are ejected regardless of light intensity.

Photosynthesis

Plants absorb photons in the visible range (mainly red at ~680 nm and blue at ~450 nm):

• Red photons (680 nm): E = 1.82 eV
• Blue photons (450 nm): E = 2.76 eV

These energies are sufficient to excite electrons in chlorophyll molecules, driving the chemical reactions that convert CO₂ and H₂O into sugars.

Vision

Human eyes detect photons in the 380-700 nm range (1.77-3.26 eV). Rod cells can respond to single photons under dark-adapted conditions. The energy of each photon triggers a conformational change in rhodopsin molecules, initiating the neural signal.

Semiconductor Band Gaps

The band gap of a semiconductor determines which photons it can absorb:

• Silicon (1.1 eV): Absorbs photons with λ < 1127 nm (visible and near-IR)
• Gallium arsenide (1.4 eV): Absorbs λ < 886 nm
• Cadmium sulfide (2.4 eV): Absorbs λ < 517 nm (green and blue)
• Diamond (5.5 eV): Absorbs λ < 225 nm (UV only, transparent to visible)

This determines the efficiency of solar cells and the emission wavelengths of LEDs.

Historical Development

Planck's Quantum Hypothesis (1900)

Max Planck introduced the concept of energy quantization to explain black-body radiation. He proposed that energy is emitted and absorbed in discrete units:

The constant h (Planck's constant) was introduced as a mathematical necessity but had no known physical meaning at the time.

Einstein's Photon Theory (1905)

Einstein went further, proposing that light itself consists of discrete particles (later called photons), each with energy E = hf. This explained the photoelectric effect and established the photon energy formula we use today. Einstein received the 1921 Nobel Prize in Physics for this work.

Bohr's Atomic Model (1913)

Niels Bohr applied the photon energy concept to atomic structure. Atoms emit photons when electrons jump between energy levels:

This explained why atoms emit light at specific wavelengths (spectral lines) rather than a continuous spectrum.

Applications

Spectroscopy

Spectroscopy uses the photon energy formula to identify substances by their absorption or emission spectra. Each element and compound has characteristic energy levels, producing unique spectral "fingerprints."

Medical Imaging

Different photon energies are used for different imaging modalities:

• Mammography X-rays: 15-25 keV (soft tissue contrast)
• Chest X-rays: 60-120 keV (bone and lung imaging)
• CT scans: 80-140 keV
• PET scans: 511 keV (gamma rays from positron annihilation)

Radiation Therapy

Cancer treatment uses high-energy photons to destroy tumor cells:

• Orthovoltage X-rays: 100-500 keV (superficial tumors)
• Linear accelerator X-rays: 4-25 MeV (deep tumors)
• Gamma knife: Cobalt-60 gamma rays at 1.17 and 1.33 MeV

Solar Cells

Photovoltaic cells convert photon energy to electrical energy. The efficiency depends on matching the band gap to the solar spectrum. Most solar energy is in the visible range (1.5-3 eV), which is why silicon (band gap 1.1 eV) works well—it absorbs most visible photons.

LEDs and Lasers

Light-emitting devices produce photons with energies equal to their material's band gap:

• Red LED (AlGaAs): ~1.9 eV → 650 nm
• Green LED (InGaN): ~2.3 eV → 540 nm
• Blue LED (InGaN): ~2.7 eV → 460 nm
• UV LED: ~3.4 eV → 365 nm

Photochemistry

Chemical reactions can be initiated or prevented based on photon energy:

• UV sterilization uses 254 nm photons (4.9 eV) to break DNA bonds in microorganisms
• Photopolymerization in dental fillings uses blue light (2.5-3 eV)
• Photodynamic therapy uses specific wavelengths to activate drugs that destroy cancer cells

Number of Photons in a Light Beam

The power of a light source equals the energy delivered per second. For monochromatic light:

Example: 60W Light Bulb

Problem: A 60-watt incandescent bulb emits mostly in the infrared. Assuming an average wavelength of 1000 nm, how many photons does it emit per second?

Solution:

1. Energy per photon: E = 1239.84/1000 = 1.24 eV = 1.99 × 10⁻¹⁹ J
2. Photons per second: N = 60 / (1.99 × 10⁻¹⁹) = 3.02 × 10²⁰ photons/second

Answer: About 300 billion billion photons per second!

Example: Green Laser Pointer

Problem: A 5 mW green laser pointer (532 nm) emits how many photons per second?

Solution:

1. Energy per photon: E = 1239.84/532 = 2.33 eV = 3.73 × 10⁻¹⁹ J
2. Photons per second: N = 0.005 / (3.73 × 10⁻¹⁹) = 1.34 × 10¹⁶ photons/second

Answer: About 13 trillion photons per second

Common Mistakes to Avoid

Mistake 1: Confusing Wavelength Units

The simplified formula E (eV) = 1239.84/λ only works when λ is in nanometers. Using meters or other units gives wrong answers.

Mistake 2: Intensity vs. Photon Energy

A brighter light has more photons per second, but each photon still has the same energy. Doubling intensity doubles photon count, not photon energy. Only changing wavelength changes photon energy.

Mistake 3: Forgetting Unit Conversions

When using E = hc/λ directly, remember:

• h must be in J·s (not eV·s)
• c must be in m/s
• λ must be in meters
• Result E will be in joules (convert to eV by dividing by 1.602 × 10⁻¹⁹)

Mistake 4: Using Classical Wave Concepts

In quantum mechanics, energy transfer happens in discrete photons, not continuously. You can't have "half a photon" of energy.

Quick Reference Formulas

Use our photon energy calculator to quickly convert between wavelength and energy in any units.

Energy-Wavelength Conversion Table for Key EM Regions

The following table provides precise photon energy values for specific wavelengths spanning the entire electromagnetic spectrum. All values are calculated using E = hc/λ with hc = 1239.84198 eV·nm. This serves as a ready reference for quickly looking up energy-wavelength relationships without recalculating.

Ionization Energies of Common Elements

A photon must have at least enough energy to overcome an atom's ionization energy in order to eject an electron. The first ionization energy is the minimum energy needed to remove the outermost electron from a neutral atom in the gas phase. The following table lists ionization energies and the corresponding maximum wavelength (threshold wavelength) capable of causing ionization, calculated using λ = 1239.84/E.

This table reveals several important physical insights. The alkali metals (cesium, potassium, sodium, lithium) have the lowest ionization energies because their outermost electron is in a new shell far from the nucleus. Cesium, with the lowest first ionization energy of any stable element at 3.89 eV, can be ionized by near-UV light just beyond the visible violet range. This makes cesium and other alkali metals ideal for photoelectric effect demonstrations and photocathode applications.

The noble gases (helium, neon, argon) have the highest ionization energies in their respective periods because their electron shells are completely filled. Helium requires 24.59 eV photons, corresponding to extreme UV radiation at just 50.4 nm. No visible or near-UV light can ionize helium atoms, which is why helium is chemically inert under ordinary conditions.

The ionization energy of hydrogen (13.60 eV, or 91.2 nm) is particularly significant in astrophysics. The corresponding wavelength of 91.2 nm is called the Lyman limit, and radiation below this wavelength is absorbed by neutral hydrogen in the interstellar medium, creating an "absorption edge" in the spectra of distant stars and galaxies.

Photon Energy Comparison Across the Spectrum

To build intuition about the vast range of photon energies, it is helpful to compare specific examples from different parts of the electromagnetic spectrum side by side. The following table highlights the enormous range of energies, from the extraordinarily weak photons of AM radio to the immensely energetic gamma rays from nuclear reactions.

The last column shows the ratio of each photon's energy to that of green visible light (550 nm, 2.254 eV). This ratio spans an astonishing range of nearly 10¹⁹ (ten billion billion) from AM radio to high-energy gamma rays. A single gamma ray photon from a cosmic burst carries roughly the same energy as 10 billion visible light photons combined.

This enormous range is precisely why different parts of the electromagnetic spectrum interact so differently with matter. Radio and microwave photons are far too weak to break chemical bonds or ionize atoms, so they pass through most materials harmlessly (though they can heat polar molecules, as in a microwave oven). Visible and UV photons carry enough energy to excite or break electronic bonds, enabling chemistry, vision, and photosynthesis. X-ray and gamma ray photons are energetic enough to ionize atoms deep within matter, making them both useful for medical imaging and hazardous to biological tissue.

Summary

The photon energy formula E = hc/λ is fundamental to modern physics:

• Energy is inversely proportional to wavelength: Short wavelengths (gamma rays, X-rays) carry high energy; long wavelengths (radio waves) carry low energy
• The constant hc = 1239.84 eV·nm enables simple calculations for visible light
• Photon energy determines what physical and chemical processes are possible (photoelectric effect, photosynthesis, ionization)
• Applications span from spectroscopy and medical imaging to solar cells and LEDs
• Visible light photons have energies of about 1.77-3.26 eV (red to violet)

Frequently Asked Questions