Chapter 26. Momentum and energy of a photon (26-7)

Question

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Question

480hXuyESyrPzdt/xvARofgN+tQMSOh3LHJ9zTD++v4xcHwmoVcmtA==
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Question

lO22s9WXxOIWyt/tFtNeLU0G7JipMTCws9/X1bgYaV6zxU8Qk7akWQ==
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Question

MrIETWh6nMy0Ebg6F+exH+eGoyUKgTCjfGVG4f3KHbe2LoHzbKqJiSlp5TM=
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Question

CTc43SGus9Fhs48wootM9pjbRUVjShhVoGI2z7nFqdeHC/pJZ0xivpSEca+eAhFY
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Review

A photon has zero mass, which is why it can travel at the speed of light. (Any object with nonzero mass would require an infinite amount of kinetic energy to reach \(v = c\), as we described in Section 25-7.) As such, we can’t apply Equation 26-3 or 26-4 directly to photons. But we can use the combination in Equation 26-5, in which mass does not appear explicitly. Setting \(v = c\) in Equation 26-5, we get the following relationship for the momentum of a photon:

(26-6) \(p = \frac{E_c}{c^2} = \frac{E}{c}\) (momentum of a photon)

From Equation 22-26 we can write \(E = hf\), and we know that for light waves \(c = f \lambda\) (Equation 22-2). If we substitute these into Equation 26-6, we get an alternative expression for the momentum of a photon:

\(p = \frac{hf}{f\lambda}\)

or, simplifying,