Answer
633 nm
Work Step by Step
Figure 28-20 shows that photons are released with energy 1.96 eV. Find the wavelength of these photons.
In this problem, the value of hc is useful. Use the result stated in Chapter 27, Problem 29, $hc=1240\;eV \cdot nm$.
Use equation 27–4, E = hf, and recall that $\lambda f = c$.
$$\lambda = \frac{c}{f}=\frac{hc}{hf}=\frac{1240\;eV \cdot nm }{1.96eV}=633nm$$
