If one photon of light is known to have energy 3.33×10^-19J, is it visible?
From the definition of the electron volt, we know 1 eV = 1.60 × 10−19 J
"E = \\frac{3.33 \\times 10^{-19}}{1.60 \\times 10^{-19}} = 2.08 \\;eV \\\\\n\n\u03bb = \\frac{hc}{E} \\\\\n\n= \\frac{1.24 \\times 10^{-6}}{2.08} = 596 \\;nm"
Visible light: λ=400−700 nm
Answer: It is visible.
Comments
Leave a comment