Calculate the magnitude of the energy of the photon associated with light of wavelength 6057.8 Å

Answer: \(3.28\times 10^{-19}\ \text{J}\) (about \(2.05\ \text{eV}\))

Explanation: Use \(E=\dfrac{hc}{\lambda}\). With \(h=6.62607015\times10^{-34}\ \text{J·s}\), \(c=2.99792458\times10^8\ \text{m/s}\), and \(\lambda=6057.8\ \text{Å}=6.0578\times10^{-7}\ \text{m}\),

\[ E=\frac{(6.62607015\times10^{-34})(2.99792458\times10^8)}{6.0578\times10^{-7}} \approx 3.28\times10^{-19}\ \text{J}. \]

Convert to electronvolts: \(E/\!e \approx 3.28\times10^{-19}\ \text{J}/1.602176634\times10^{-19}\ \text{J/eV}\approx 2.05\ \text{eV}.\)

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