A photon of ultraviolet radiation has a frequency of $1.5 \times 10^{15} \text{ Hz}$ - AQA - A-Level Physics - Question 11 - 2019 - Paper 1

To find the momentum $p$ of a photon, we can use the formula:

$p = \frac{E}{c}$

where:

• $E$ is the energy of the photon
• $c$ is the speed of light, approximately $3.0 \times 10^8 \text{ m s}^{-1}$

The energy of the photon can be calculated using the formula:

$E = h f$

where:

• $h$ is Planck's constant, approximately $6.63 \times 10^{-34} \text{ J s}$
• $f$ is the frequency of the photon

Substituting the values:

$E = (6.63 \times 10^{-34} \text{ J s}) \times (1.5 \times 10^{15} \text{ Hz}) = 9.945 \times 10^{-19} \text{ J}$

Now substituting for $p$:

$p = \frac{9.945 \times 10^{-19} \text{ J}}{3.0 \times 10^{8} \text{ m s}^{-1}} = 3.315 \times 10^{-27} \text{ kg m s}^{-1}$

Converting to scientific notation, we have:

$p \approx 3.3 \times 10^{-27} \text{ kg m s}^{-1}$

Thus, the answer that corresponds with the options given is:

C. $3.3 \times 10^{-27} \text{ kg m s}^{-1}$.