Question

Calculate the wavelength of the radiation emitted, when an excited atom of energy band gap 12.43 keV comes back to ground state.

Answer 1

Solution.

$\Delta E = \frac{h \times c}{\lambda}$

$\lambda = \frac{h \times c}{\Delta E}$

$\lambda = \frac{6.63*10^{-34} \times 3*10^8}{1.99*10^{-15}} = 0.01 \ nm$

Answer:

$\lambda = 0.01 \ nm$