Question

For light of wavelength $\lambda$ in nanometer, the photon energy $hf$ in electron-volt is

• A. $\frac{1240}{\lambda }$
• B. $\frac{1200}{\lambda }$
• C. $\frac{\lambda }{1240}$
• D. $\frac{1360}{\lambda }$

Answer 1

The correct option is A $\frac{1240}{\lambda }$

Photon energy is given by:

$E=\frac{hc}{\lambda }$

$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{\lambda \times {10}^{-9}}J$

$\approx \frac{2\times {10}^{-16}}{\lambda }\times 6.2\times {10}^{18}eV$

$=\frac{1240}{\lambda }eV$