Question

If \( \lambda \) be the wavelength of any electromagnetic radiation, the de-Broglie wavelength of its quantum (photon) is

• A. \( \frac{\lambda}{4} \)
• B. \( \lambda \)
• C. \( \frac{\lambda}{2} \)
• D. \( 2\lambda \)
• E. \( \frac{3\lambda}{4} \)

Hint:

Photon already follows de-Broglie relation.

Answer 1

The Correct Option is B

Step 1: Understanding de-Broglie Wavelength and Photon Momentum:
The de-Broglie hypothesis proposes that all matter exhibits wave-like behavior. The de-Broglie wavelength (\(\lambda_{dB}\)) associated with any particle is given by the relation:
\[ \lambda_{dB} = \frac{h}{p} \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle.
A photon is the quantum, or particle, of electromagnetic radiation. We need to find its de-Broglie wavelength.
Step 2: Key Formula or Approach:
First, we need to determine the momentum (\(p\)) of a photon. The energy (\(E\)) of a photon is given by two fundamental relations:
\(E = hf = \frac{hc}{\lambda}\) (from the wave nature of light)
\(E = pc\) (from the theory of relativity for a massless particle)
Step 3: Detailed Calculation:
By equating the two expressions for the energy of a photon, we can find its momentum:
\[ pc = \frac{hc}{\lambda} \] Dividing both sides by \(c\) gives the momentum of the photon:
\[ p = \frac{h}{\lambda} \] Now, we substitute this expression for momentum into the de-Broglie wavelength formula:
\[ \lambda_{dB} = \frac{h}{p} = \frac{h}{(h/\lambda)} \] \[ \lambda_{dB} = h \times \frac{\lambda}{h} = \lambda \] Step 4: Final Answer:
The de-Broglie wavelength of a photon is exactly equal to the wavelength of the electromagnetic radiation it constitutes.