Question

If one photon of light is known to have energy 3.33×10^-19J, is it visible?

Answer 1

From the definition of the electron volt, we know 1 eV = 1.60 × 10⁻¹⁹ J

$E = \frac{3.33 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.08 \;eV \\ λ = \frac{hc}{E} \\ = \frac{1.24 \times 10^{-6}}{2.08} = 596 \;nm$

Visible light: λ=400−700 nm

Answer: It is visible.