1 \begin{problem*}{28.14}
2 Calculate the energy and momentum of a photon of wavelength $700\U{nm}$.
3 \end{problem*} % problem 28.14
6 You should be familiar with these equations by now (after our time
7 with the Bohr atom and relativity).
8 The energy is given by (Eq's 28.3 and 11.15)
10 E = \frac{hc}{\lambda} = \frac{1240\U{eV$\cdot$nm}}{700\U{nm}}
11 = \ans{1.77\U{eV}} \;.
13 For momentum you can either use the relativistic energy-momentum
16 E^2 &= p^2c^2 + m_0^2c^4 \\
17 E_\text{photon} &= p c \\
18 p &= \frac{E}{c} = \ans{1.77\U{eV/$c$}} = \ans{9.46\E{-28}\U{kg$\cdot$m/s}}
20 or the de Broglie formula (Eq.~28.10)
22 p &= \frac{h}{\lambda} = \frac{6.63\E{-34}\U{J$\cdot$s}}{700\U{nm}}
23 = \ans{9.46\E{-28}\U{kg$\cdot$m/s}} \;.