Energy of a photon of wavelength 5890Åemitted by sodium vapour lamp is

Correct option is C

​$$\textbf{Given:} \\\bullet \text{Wavelength, } \lambda = 5890\, \text{\AA} = 5890 \times 10^{-10}\, \text{m} \\\bullet \text{Speed of light, } c = 3 \times 10^8\, \text{m/s} \\\bullet \text{Planck's constant, } h = 6.626 \times 10^{-34}\, \text{J} \cdot \text{s} \\\bullet 1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J}\\[10pt]\textbf{Formula } \\E = \frac{hc}{\lambda}\\[10pt]E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5890 \times 10^{-10}} \\E \approx 3.3748 \times 10^{-19}\, \text{J}\\[10pt]\text{Now convert joules to eV:} \\E = \frac{3.3748 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.1\, \text{eV}$$​​