The momentum of a photon of wavelength 6626 nm will be — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - de Broglie Relation for Photons The momentum $P$ of a photon is given by: $$P = \frac{h}{\lambda}$$ Where $h = 6.626 \times 10^{-34} \text{ J s}$ (Planck's constant) and $\lambda$ is the wavelength in meters. --- ### Step 2 - Convert Wavelength to SI Units Given $\lambda = 6626 \text{ nm}$: $$\lambda = 6626 \times 10^{-9} \text{ m} = 6.626 \times 10^{-6} \text{ m}$$ --- ### Step 3 - Calculate Momentum $$P = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{6.626 \times 10^{-6}}$$ The coefficients cancel: $$P = 1 \times 10^{-34-(-6)} = 1 \times 10^{-28} \text{ kg m s}^{-1}$$ $$P = \boxed{10^{-28} \text{ kg m s}^{-1}}$$ --- ### Step 4 - Analysis of Options * **Option (A) $10^{-28}$ kg ms$^{-1}$:** Correct — matches our calculation. * **Option (B) $10^{-25}$ kg ms$^{-1}$:** Incorrect — error in exponent arithmetic. * **Option (C) $10^{-31}$ kg ms$^{-1}$:** Incorrect — wrong unit conversion of wavelength. * **Option (D) zero:** Incorrect — a photon with finite wavelength always has non-zero momentum.