Question

In the mercury spectrum one can find a strong emission line with the wavelength

576,960 nm. This corresponds to the energy difference between two energy levels in the mercury atom. Determine this energy difference expressed in electron volts.

Answer 1

According to the Planck's formula, the energy correspondes to the wavelength $\lambda = 576,960 nm$ (that is, the energy difference between two energy levels in the mercury atom) will be:

$E = h\cdot \dfrac{c}{\lambda}$,

where $h = 4.14\cdot 10^{-15}\space eV\cdot s$ is the Planck's constant and $c = 3\cdot 10^8 m/s$ is the speed of light. Thus:

$E = 4.14\cdot 10^{-15}\space \cdot \dfrac{3\cdot 10^8}{576.960 \cdot 10^{-9}} \approx 2.15 eV$ .

Answer. E = 2.15 eV.