Question

The momentum of a photon associated with a microwave of wavelength 4.00 cm is:

• A. \(1.66 \times 10^{-32}\) kg ms⁻¹
• B. \(1.83 \times 10^{-34}\) kg ms⁻¹
• C. \(2.05 \times 10^{-34}\) kg ms⁻¹
• D. \(1.66 \times 10^{-34}\) kg ms⁻¹

Hint:

Always double-check units in Dual Nature problems; wavelength must be in meters to match the SI units of Planck's constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Photons carry momentum despite having no rest mass. The de Broglie relation links the particle-like property (momentum) to the wave-like property (wavelength).

Step 2: Key Formula or Approach:
Momentum \(p = \frac{h}{\lambda}\), where Planck's constant \(h \approx 6.63 \times 10^{-34}\) J s.

Step 3: Detailed Explanation:
First, convert the wavelength to SI units: \(\lambda = 4.00\) cm \(= 0.04\) m. Now, calculate the momentum: \[ p = \frac{6.63 \times 10^{-34}}{0.04} \] \[ p = 1.6575 \times 10^{-32} \text{ kg ms}^{-1} \] Rounding to significant figures gives \(1.66 \times 10^{-32}\) kg ms⁻¹.

Step 4: Final Answer:
The momentum of the photon is \(1.66 \times 10^{-32}\) kg ms⁻¹.