The energy ($E$) of a photon is related to its wavelength ($\lambda$) by the Planck's equation: $E = \frac{hc}{\lambda}$

Rearranging the formula to solve for wavelength: $\lambda = \frac{hc}{E}$

Given: Planck's constant, $h = 6.6 \times 10^{-34} \text{ J s}$ Speed of light, $c = 3 \times 10^8 \text{ m/s}$

• Energy, $E = 3.03 \times 10^{-19} \text{ J}$

Substituting the values: $\lambda = \frac{(6.6 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{3.03 \times 10^{-19} \text{ J}}$ $\lambda = \frac{19.8 \times 10^{-26}}{3.03 \times 10^{-19}} \text{ m}$ $\lambda \approx 6.534 \times 10^{-7} \text{ m}$

Converting to nanometers ($1 \text{ nm} = 10^{-9} \text{ m}$): $\lambda = 6.534 \times 10^{-7} \times 10^9 \text{ nm}$ $\lambda = 653.4 \text{ nm}$

This matches Option 4.