A photon of wavelength $300\,nm$ has approximately what energy in eV? (Use $hc = 1240\,eV\,nm$.)
A$2.07\,eV$
B$8.26\,eV$
C$4.13\,eV$
D$0.413\,eV$
Answer & Solution
Correct answer: C. $4.13\,eV$
$E = hc/\lambda = 1240/300 \approx 4.13\,eV$. The shortcut $hc = 1240\,eV\,nm$ is worth memorising — divides wavelength in nm directly to give energy in eV.
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