Momentum and energy of a photon (26-7)

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Question

Wavelength of the photon

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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,