Answer
$f_{min}\approx 2\times10^{13}Hz$
$\lambda_{max}\approx 1 \times10^{-5}m$
Work Step by Step
Use equation 27–4, E = hf, and recall that $\lambda f = c$.
$$E_{min}=hf_{min}$$
$$f_{min}=\frac{E_{min}}{h}=\frac{(0.1eV)(1.60\times10^{-19}J/eV)}{6.63\times10^{-34}J \cdot s }$$
$$=2.41\times10^{13}Hz\approx 2\times10^{13}Hz$$
$$\lambda_{max}=\frac{c}{f_{min}}=\frac{3.00\times10^8m/s}{2.41\times10^{13}Hz }\approx 1 \times10^{-5}m$$
