Answer
See answers.
Work Step by Step
Use equation 27–4, E = hf, to find the energy of a photon. For light, we also know that $f=c/\lambda$. The longest wavelength photon has the lowest energy.
$$E_{low}=hf_{low}=\frac{hc}{\lambda_{high}}=\frac{(6.63\times10^{-34}J\cdot s)(3.00\times10^8 m/s)}{750\times10^{-9}m}$$
$$E_{low}=2.7\times10^{-19}J=1.7eV$$
Now calculate the energy of the most energetic photons in the visible spectrum.
$$E_{high}=hf_{low}=\frac{hc}{\lambda_{low}}=\frac{(6.63\times10^{-34}J\cdot s)(3.00\times10^8 m/s)}{400\times10^{-9}m}$$
$$E_{high}=5.0\times10^{-19}J=3.1eV$$
Put all the facts together about photons in the visible spectrum.
$$2.7\times10^{-19}J\lt E \lt 5.0\times10^{-19}J$$
$$1.7 eV\lt E \lt 3.1eV$$
