Neils Bohr numbered the energy levels (n) of hydrogen, with level 1 (n=1) being the ground state, level 2 being the first excited state, and so on. Remember that there is a maximum energy that each electron can have and still be part of its atom. Beyond that energy, the electron is no longer bound to the nucleus of the atom and it is considered to be ionized. In that case n approaches infinity.
The equation for determining the energy of any state (the nth) is as follows:
E = -13.6/n2 eV
Because the energy is so small, the energy is measured in electron-volts, designated as “eV”. 1 eV = 1.6 × 10-19 J.
Answer the following questions:
For the first excited state, n=2. Using this in the above equation gives E = -3.40 eV
The energy of the photon is found by computing the difference
in the energies of the fourth (n=4) and second (n=2) levels
E = -13.6/42 – (-13.6/22)
E = -0.85 + 3.40
E = 2.55 eV
We represent a free electron by assigning it an infinite n. Hence, its energy is zero.
The energy of the photon emitted by a free electron captured to the n=2 level is thus
E = 0 – (-13.6/22) = 3.4 eV
From the Calculation Investigation, we learned that energy and wavelength are related through E = h c / l.
We can solve this for the wavelength, l = h c / E. where h = 6.626 × 10-34J-s, and c = 3 × 108 m/s
We convert our energy E= 2.55 ev into Joules using 1 eV = 1.6×10-19 J. This gives an energy of E = 4.08 × 10-19 J
We then find a wavelength of
l = ((6.626 × 10-34) x (3 × 108)) / (4.08 × 10-19)
l = 4.87 × 10-7 mOr, using 1 nm = 1 × 10-9 m,
l = 487 nm.
The transition is the bright blue line, just to the left of the center.