Q: What working with light and X- rays, there is a useful relation between the energy of a photon in electron volts (eV) and the wavelength of the photon in angstrom (A°). Suppose the wavelength of a photon is λ A° Then energy of the photon is

Sol: E = hv = hc/λ

Here , wavelength = λ × 10^{-10} m ; h = 6.62 × 10^{-34} Js , c = 3 × 10^{8} m/s

$\large E = \frac{6.62 \times 10^{-34}\times 3 \times 10^8}{\lambda \times 10^{-10}} $

$\large E = \frac{6.62 \times 10^{-34}\times 3 \times 10^8}{\lambda \times 10^{-10}\times 1.6 \times 10^{-19}} eV $

$\large E = \frac{12400}{\lambda(A^o)} eV$

Note : ( λ is taken in A° and 12400 in A° eV )