Question

A calcium atom drops from 5.320 eV above the ground state to 4.130 eV above the ground state. What is the wavelength of the photon emitted?

Answer 1

$\Delta E=h\nu=hc/\lambda\to \lambda=hc/\Delta E=$

$=6.626\cdot10^{-34}\cdot3\cdot10^{8}/((5.32-4.13)\cdot 1.6\cdot10^{-19})=$

$=1.044\cdot10^{-6}\ (m)=1.044\ \mu m$ . Answer