When an electron in a certain excited state in a Helium atom relaxes to a lower

Question

When an electron in a certain excited state in a Helium atom relaxes to a lower energy state, it can emit a photon. Two types of photons can be emitted from this state: a photon with wavelength 58.4 nm, or a photon with wavelength 2.06 µm. Each of these corresponds to one transition from this same excited state to a lower-energy state. a) From the energies of these two wavelengths, construct an energy-level diagram of the three energy levels involved. Show the two energy transitions by downward arrows starting from the initial state and pointing to the final state. b) Calculate the energy difference between the two lowest states

Explanation / Answer

Wavelength of the first photon emitted ----> 5.84 x10-8m Emission to level 1

Wavelength of the second photon emitted ------> 2.06 X10-6m emission to level 2

Lower is the Wavelength, Higher is the Energy.

therefore Level 1 lies above level 2 and the excited state is above these levels.

Energy difference = Level1 - Level2

= 5.84 X 10-8 - 2.06 X 10-6

= 10-6(584-2.06)

= 581.94 X 10-6m

= 581.94 um (micro metre)

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