Question

The momentum (in kg·m/s) of a photon of frequency \( 6.0 \times 10^{14} \) Hz is:

• A. \( 6.63 \times 10^{-25} \)
• B. \( 1.326 \times 10^{-27} \)
• C. \( 2.652 \times 10^{-26} \)
• D. \( 3.978 \times 10^{-24} \)

Hint:

To find photon momentum, use \( p = \frac{hf}{c} \). Use powers of ten carefully and simplify step-by-step for accurate results.

Answer 1

The Correct Option is B

The momentum \( p \) of a photon is given by: \[ p = \frac{E}{c} = \frac{hf}{c} \] Where: \( h = 6.63 \times 10^{-34} \, \text{J·s} \) (Planck’s constant) \( f = 6.0 \times 10^{14} \, \text{Hz} \) (frequency) \( c = 3.0 \times 10^8 \, \text{m/s} \) (speed of light) Substitute the values: \[ p = \frac{6.63 \times 10^{-34} \times 6.0 \times 10^{14}}{3.0 \times 10^8} \] First, calculate the numerator: \[ 6.63 \times 6.0 = 39.78,\quad \text{and powers: } 10^{-34} \times 10^{14} = 10^{-20} \] So, numerator = \( 39.78 \times 10^{-20} \) Now divide by \( 3.0 \times 10^8 \): \[ p = \frac{39.78 \times 10^{-20}}{3.0 \times 10^8} = 13.26 \times 10^{-28} = 1.326 \times 10^{-27} \, \text{kg·m/s} \]