Q.Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Concept: De Broglie Wavelength — the idea that every particle with momentum has an associated wavelength $\lambda = h/p$.

For a photon of electromagnetic radiation:

1. The photon energy is $E = h\nu$, and its momentum is $p = \frac{E}{c}$ (from Maxwell's theory).
2. Substitute $E = h\nu$ into the momentum: $p = \frac{h\nu}{c}$.
3. Since $\nu = c/\lambda$, we get $p = \frac{h}{\lambda}$. …

Why This Works — The Core Idea

The de Broglie hypothesis extends wave-particle duality to all matter: every moving particle has an associated wavelength. For a photon — the quantum of light — this is not a separate concept. Light already behaves as a wave with wavelength $\lambda$, and as a particle with energy $E = h\nu$ and momentum $p = h/\lambda$. The de Broglie relation simply recovers the same wavelength from the particle’s momentum. So the two are not just equal; they are the same physical quantity expressed from different starting points.

Step-by-Step Derivation

1. Recall the de Broglie wavelength formula

For any particle with momentum $p$, the de Broglie wavelength is:

where $h$ is Planck’s constant. This is the fundamental relation.

2. Find the momentum of a photon

A photon has energy $E = h\nu$, where $\nu$ is the frequency of the electromagnetic wave. For light, the wave speed is $c$, so $\nu = c/\lambda$. Thus:

From special relativity, a photon’s energy and momentum are related by $E = pc$ (since the photon is massless). Therefore:

3. Substitute into the de Broglie relation

Plug $p = h/\lambda$ into $\lambda_{\text{dB}} = h/p$:

That’s it. The de Broglie wavelength of the photon is exactly the wavelength of the electromagnetic radiation. …

Method: Direct Substitution Using the de Broglie Relation and Photon Momentum

The idea is simple: the de Broglie wavelength formula applies to any particle with momentum — and a photon is a particle with momentum. So we just plug the photon's momentum into the de Broglie relation and see what comes out.

Steps

Step 1 — Write the de Broglie wavelength formula

For any particle, the de Broglie wavelength $\lambda$ is given by:

where $h$ is Planck's constant and $p$ is the linear momentum of the particle.

Step 2 — Write the momentum of a photon

A photon of electromagnetic radiation of frequency $\nu$ (or wavelength $\lambda_{\text{EM}}$) carries energy $E = h\nu$. From special relativity, the momentum of a massless particle (like a photon) is:

Substitute $E = h\nu$:

Step 3 — Substitute photon momentum into de Broglie relation

Put this $p$ into $\lambda = h/p$:

Step 4 — Relate $c/\nu$ to the electromagnetic wavelength …

Students often treat this as a straightforward substitution, but the logic has subtle traps. Here are the most common mistakes and how to avoid each.

Mistake 1: Using the wrong expression for momentum of a photon

Many students write $p = mv$ and then try to plug in $m = 0$ for a photon. That leads to $p = 0$, which makes the de Broglie wavelength $\lambda = h/p$ undefined — and the proof collapses.

How to avoid: Remember that for a photon, momentum is not $mv$. The correct relation comes from electromagnetic theory: $p = \frac{E}{c}$. Since $E = h\nu$ and $\nu = c/\lambda$, you get $p = \frac{h}{\lambda}$. This is the only valid starting point for a photon.

Mistake 2: Confusing the two different $\lambda$ symbols

In the question, "wavelength of electromagnetic radiation" (call it $\lambda_{\text{EM}}$) and "de Broglie wavelength" (call it $\lambda_{\text{dB}}$) are conceptually distinct. Students often assume they are the same symbol and skip the derivation, writing $\lambda = h/p$ and then $p = h/\lambda$ — which is circular.

How to avoid: Write them separately at the start. Let $\lambda_{\text{EM}}$ be the radiation wavelength from $c = \nu \lambda_{\text{EM}}$. Let $\lambda_{\text{dB}}$ be the de Broglie wavelength $h/p$. Your goal is to show $\lambda_{\text{EM}} = \lambda_{\text{dB}}$.

Mistake 3: Forgetting that $E = h\nu$ applies to the photon

Some students try to use $E = mc^2$ for the photon, which again hits the $m=0$ problem. Others use $E = \frac{1}{2}mv^2$, which is completely wrong for a photon.

How to avoid: Anchor yourself: a photon's energy is only given by $E = h\nu$ (or $E = hc/\lambda_{\text{EM}}$). Never substitute mass-based formulas for a photon.

Mistake 4: Mixing up $c$ and $v$ in the de Broglie relation

For a massive particle, de Broglie wavelength is $\lambda = h/(mv)$. Some students carry this over and write $p = m c$ for a photon, which is technically correct only if you treat $m$ as relativistic mass — but that introduces unnecessary confusion and is not standard in most Indian board syllabi.

How to avoid: Stick to the photon-specific momentum $p = E/c$. It's cleaner and exam-safe. If your syllabus uses relativistic mass, mention it only as an alternative, but lead with $p = E/c$.

Mistake 5: Not showing the final equality clearly

After deriving $\lambda_{\text{dB}} = h/(E/c) = hc/E$ and $E = hc/\lambda_{\text{EM}}$, students sometimes stop without substituting, or they write $\lambda_{\text{dB}} = \lambda_{\text{EM}}$ without showing the cancellation.

How to avoid: Write the full chain:

$\lambda_{\text{dB}} = \frac{h}{p} = \frac{h}{E/c} = \frac{hc}{E}$ …