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 photon momentum \( p \) is calculated using the formula:\[p = \frac{E}{c} = \frac{hf}{c}\]Given constants:\( h = 6.63 \times 10^{-34} \, \text{J·s} \) (Planck’s constant)\( f = 6.0 \times 10^{14} \, \text{Hz} \) (frequency)\( c = 3.0 \times 10^8 \, \text{m/s} \) (speed of light)Substituting the values:\[p = \frac{6.63 \times 10^{-34} \times 6.0 \times 10^{14}}{3.0 \times 10^8}\]Numerator calculation:\[6.63 \times 6.0 = 39.78,\quad 10^{-34} \times 10^{14} = 10^{-20}\]Numerator = \( 39.78 \times 10^{-20} \)Division:\[p = \frac{39.78 \times 10^{-20}}{3.0 \times 10^8} = 13.26 \times 10^{-28} = 1.326 \times 10^{-27} \, \text{kg·m/s}\]