What is the momentum of a photon of light that has a frequency of $2.8 \times 10 ^ { 14 } \mathrm {~Hz}$ ? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$
A) $1.5 \times 10 ^ { - 28 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $6.2 \times 10 ^ { - 28 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $4.4 \times 10 ^ { - 28 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $6.3 \times 10 ^ { - 28 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $2.1 \times 10 ^{- 28} \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Question

Quizplus · Accepted Answer

The momentum $p$ of a photon can be calculated using the formula $p = \frac{h}{\lambda}$, where $h$ is Planck's constant and $\lambda$ is the wavelength of the photon. The wavelength $\lambda$ can be found from the frequency $f$ using the relation $\lambda = \frac{c}{f}$, where $c$ is the speed of light. Substituting the given values, $f = 2.8 \times 10^{14} \, \text{Hz}$, $c = 3.00 \times 10^8 \, \text{m/s}$, and $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$, we find $\lambda = \frac{3.00 \times 10^8}{2.8 \times 10^{14}}$ and then calculate $p = \frac{6.626 \times 10^{-34}}{\lambda}$, resulting in $p \approx 6.2 \times 10^{-28} \, \text{kg} \cdot \text{m/s}$.

What Is the Momentum of a Photon of Light That 2.8×1014 Hz2.8 \times 10 ^ { 14 } \mathrm {~Hz}2.8×1014 Hz

What Is the Momentum of a Photon of Light That $2.8 \times 10 ^ { 14 } \mathrm {~Hz}$