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

Question

Quizplus · Accepted Answer

The momentum of a photon is given by the equation p = h/λ, where h is Planck's constant and λ is the wavelength of the photon. The wavelength of a photon is inversely proportional to its frequency, so the higher the frequency, the shorter the wavelength. In this case, the frequency of the photon is 2.8 × $10^{14}$ Hz, so the wavelength is:$$\lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m/s}}{2.8 \times 10^{14} \text{ Hz}} = 1.07 \times 10^{-6} \text{ m}$$Now we can calculate the momentum of the photon:$$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34} \text{ J s}}{1.07 \times 10^{-6} \text{ m}} = 6.2 \times 10^{-28} \text{ kg m/s}$$Therefore, the correct answer is E) 6.2 × $10^{ - 28}$ kg∙m/s.

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