Answer to Question #115302 in Atomic and Nuclear Physics for Vidurjah Perananthan

Question #115302
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.
1
Expert's answer
2020-05-15T08:43:38-0400

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

E=hcλE = h\cdot \dfrac{c}{\lambda},

where h=4.141015 eVsh = 4.14\cdot 10^{-15}\space eV\cdot s is the Planck's constant and c=3108m/sc = 3\cdot 10^8 m/s is the speed of light. Thus:

E=4.141015 3108576.9601092.15eVE = 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.


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