6.4 Bohr’s model of the hydrogen atom  (Page 6/37)

The hydrogen atom may have other energies that are higher than the ground state. These higher energy states are known as excited energy states of a hydrogen atom .

There is only one ground state, but there are infinitely many excited states because there are infinitely many values of n in [link] . We say that the electron is in the “first exited state” when its energy is $E_2$ (when $n=2$ ), the second excited state when its energy is $E_3$ (when $n=3$ ) and, in general, in the n th exited state when its energy is $E_{n+1}.$ There is no highest-of-all excited state; however, there is a limit to the sequence of excited states. If we keep increasing n in [link] , we find that the limit is $\text{−}\underset{n\to \infty }{\text{lim}}{E}_0\text{/}{n}^2=0.$ In this limit, the electron is no longer bound to the nucleus but becomes a free electron. An electron remains bound in the hydrogen atom as long as its energy is negative. An electron that orbits the nucleus in the first Bohr orbit, closest to the nucleus, is in the ground state, where its energy has the smallest value. In the ground state, the electron is most strongly bound to the nucleus and its energy is given by [link] . If we want to remove this electron from the atom, we must supply it with enough energy, $E_{\infty },$ to at least balance out its ground state energy $E_1:$

The energy that is needed to remove the electron from the atom is called the ionization energy    . The ionization energy $E_{\infty }$ that is needed to remove the electron from the first Bohr orbit is called the ionization limit of the hydrogen atom    . The ionization limit in [link] that we obtain in Bohr’s model agrees with experimental value.

Spectral emission lines of hydrogen

To obtain the wavelengths of the emitted radiation when an electron makes a transition from the n th orbit to the m th orbit, we use the second of Bohr’s quantization conditions and [link] for energies. The emission of energy from the atom can occur only when an electron makes a transition from an excited state to a lower-energy state. In the course of such a transition, the emitted photon carries away the difference of energies between the states involved in the transition. The transition cannot go in the other direction because the energy of a photon cannot be negative, which means that for emission we must have $E_n>E_m$ and $n>m.$ Therefore, the third of Bohr’s postulates gives

Now we express the photon’s energy in terms of its wavelength, $hf=hc\text{/}\lambda ,$ and divide both sides of [link] by $hc.$ The result is

The value of the constant in this equation is