Q.The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = h\nu$ (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

$E=h\nu=hc/\lambda$, and $hc\approx1240$ eV*nm, so $E(\text{eV})=1240/\lambda(\text{nm})$.

Using one representative wavelength per band: radio (~1 m) $\approx1.24\times10^{-6}$ eV; microwave (~1 cm) $\approx1.24\times10^{-4}$ eV; infrared (~10 $\mu$m) $\approx0.124$ eV; visible (400-700 nm) $\approx1.8$-$3.1$ eV; ultraviolet (~100 nm) $\approx12.4$ eV; X-ray (~0.1 nm) $\approx1.24\times10^{4}$ eV; gamma ray (~1 pm) $\approx1.24\times10^{6}$ eV. …

The formula

with $h=6.63\times10^{-34}$ J*s, $c=3\times10^{8}$ m/s. Converting to electronvolts ($1$ eV $=1.6\times10^{-19}$ J) gives the handy shortcut

Photon energy across the spectrum (one representative wavelength per band)

Sample calculations: radio $E=1240/(1\times10^{9})=1.24\times10^{-6}$ eV; gamma ray $E=1240/(0.001)=1.24\times10^{6}$ eV.

How photon energy relates to the source

Each energy scale matches the energy scale of the physical process that produces it:

• Radio/microwave ($\mu$eV-meV): energies of oscillating currents in circuits/antennas or molecular rotations -- very small, so these come from low-energy motions.
• Infrared (~0.1 eV): matches the spacing of molecular vibrational energy levels and thermal (blackbody) emission from warm objects. …

Method: Direct Photon Energy Calculation Using $E = h\nu$

Concept: The energy of a single photon is directly proportional to its frequency. The constant of proportionality is Planck's constant $h$.

Steps

1. Recall the formula The energy of one photon is:

where:

• $E$ = energy (in joules, J)
• $h = 6.626 \times 10^{-34} \ \text{J·s}$ (Planck's constant)
• $\nu$ = frequency of radiation (in Hz, or s$^{-1}$)

2. Convert energy from joules to electronvolts (eV) Since 1 eV = $1.602 \times 10^{-19}$ J, the energy in eV is:

3. Use the relation between frequency and wavelength If wavelength $\lambda$ (in metres) is given instead of frequency:

where $c = 3.00 \times 10^8$ m/s (speed of light).

Then:

and in eV:

4. Plug in values for each spectral region For example, for visible light (say $\lambda \approx 500$ nm = $5.00 \times 10^{-7}$ m):

5. Interpret the scale
   • Radio waves → very low energy (micro-eV to meV)
   • Microwaves → low energy (meV range)
   • Infrared → moderate energy (meV to eV)
   • Visible light → ~1.5–3 eV
   • Ultraviolet → higher energy (3–100 eV) …

Here are the common mistakes students make when solving photon energy problems (like the one from the NCERT Electromagnetic Spectrum chapter), along with how to avoid each.

1. Forgetting to Convert Frequency to Hz (or Wavelength to Meters)

The Mistake:

Students plug in frequency given in MHz, GHz, or wavelength in nm, Å, or cm directly into $E = h\nu$ or $E = \frac{hc}{\lambda}$ without converting to Hz (s⁻¹) or meters (m).

Why it’s wrong:

Planck’s constant $h = 6.63 \times 10^{-34} \, \text{J s}$ and $c = 3 \times 10^8 \, \text{m/s}$ are in SI units. Using non-SI units gives a completely wrong numerical value.

How to avoid:

• Always write the given value with its unit.
• Convert:
  • $1 \, \text{MHz} = 10^6 \, \text{Hz}$
  • $1 \, \text{nm} = 10^{-9} \, \text{m}$
  • $1 \, \text{Å} = 10^{-10} \, \text{m}$
• Check: If your answer is off by a factor of $10^9$ or $10^{15}$, you likely missed a conversion.

2. Using the Wrong Formula for Energy

The Mistake:

Using $E = h\nu$ when wavelength is given, or using $E = \frac{hc}{\lambda}$ when frequency is given — but then forgetting to use $c$ correctly or mixing up $\nu$ and $\lambda$.

Why it’s wrong:

The two formulas are equivalent only if you use the correct relation $c = \nu \lambda$. Mixing them without substitution gives nonsense.

How to avoid:

• If frequency is given → use $E = h\nu$
• If wavelength is given → use $E = \frac{hc}{\lambda}$
• Never plug a wavelength into $E = h\nu$ — you’ll get energy in Joules that is off by a factor of $10^8$.

3. Incorrect Conversion from Joules to eV

The Mistake:

Using $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$ but dividing instead of multiplying (or vice versa).

Why it’s wrong:

1 eV is a smaller unit than 1 J. So a number in Joules should be divided by $1.6 \times 10^{-19}$ to get eV. Doing the opposite gives a ridiculously tiny number.

How to avoid:

• Remember: 1 eV = $1.6 \times 10^{-19}$ J
• To convert J → eV: divide by $1.6 \times 10^{-19}$
• Quick sanity check: Visible light photons are ~2–3 eV. If your answer is $10^{-19}$ eV or $10^{20}$ eV, you’ve swapped.

4. Forgetting to Use $h$ and $c$ with Correct Units

The Mistake:

Using $h = 6.63 \times 10^{-34} \, \text{J s}$ but then using $c = 3 \times 10^8 \, \text{m/s}$ and wavelength in cm — or using $h$ in eV·s without converting.

Why it’s wrong:

$h$ in J·s and $c$ in m/s require length in meters. If you use cm, the energy will be off by $10^2$.

How to avoid:

• Stick to SI units for $h$, $c$, $\lambda$, $\nu$.
• If you want eV directly, use the convenient constant:

Then $E (\text{eV}) = \frac{1240}{\lambda (\text{nm})}$ — this avoids unit errors entirely.

5. Misinterpreting the “Scale” of Photon Energies

The Mistake:

After calculating energies, students fail to connect the order of magnitude to the source (e.g., radio waves have tiny eV, gamma rays have huge MeV). They just list numbers without insight.

Why it’s wrong:

The question explicitly asks: “In what way are the different scales of photon energies related to the sources?” — this is a conceptual link.

How to avoid:

• Compare orders of magnitude:
  • Radio: $\sim 10^{-6}$ eV (electronic circuits)
  • Microwave: $\sim 10^{-3}$ eV (molecular rotation)
  • Visible: $\sim 1$–$3$ eV (electronic transitions in atoms)
  • X-ray: $\sim 10^3$–$10^5$ eV (inner shell transitions) …