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(C) The momentum of a photon is given by $p = \frac{h}{\lambda}$.Given $\lambda = 6630 \mathring{A} = 6630 	imes 10^{-10} \ m$ and $h = 6.63 	imes 1

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$1 	imes 10^{-27} \ kg \cdot m/s$

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(C) The momentum of a photon is given by $p = \frac{h}{\lambda}$.Given $\lambda = 6630 \mathring{A} = 6630 	imes 10^{-10} \ m$ and $h = 6.63 	imes 10^{-34} \ J \cdot s$.For a totally reflecting surface,the change in momentum is $\Delta p = p_{final} - p_{initial} = -p - p = -2p$ (magnitude is $2p$).However,the question asks for the momentum delivered by the photon itself,which is $p = \frac{h}{\lambda}$.$p = \frac{6.63 	imes 10^{-34} \ J \cdot s}{6630 	imes 10^{-10} \ m} = 1 	imes 10^{-27} \ kg \cdot m/s$.

$A$ photon of wavelength $6630 \mathring{A}$ is incident on a totally reflecting surface. The momentum delivered by the photon is equal to

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