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What is Ionization Energy and Ionization Potential? What was the Reason for the Failure of the Bohr Model?
Ionisation energy: The minimum amount of energy required to ionise an atom of an element at its ground state is called ionisation energy of that element.
Example: The ground state energy of hydrogen atom = -13.6 eV. In a H+ ion, the only electron of the corresponding hydrogen atom has been removed. In this state, the electron is no longer attracted by the atom, i.e., its potential energy becomes zero.
Again, the condition for minimum energy to be possessed by the electron is that its kinetic energy is zero; as a result its total energy = kinetic energy + potential energy = 0. Naturally, if the electron is brought from -13.6 eV energy state to zero energy state, the atom can be converted into an ion. So, the minimum amount of energy supplied = 0 – (-13.6) = 13.6 eV.
Hence, the ionisation energy of hydrogen 13.6 eV.
Ionisation potential: The minimum potential to be applied to an atom of an element in its ground state to convert it into a positive ion, is called the ionisation potential of that element.
Example: According to the definition, if 1 V potential is applied to an electron having charge – e, the gain in energy of the electron = 1 eV.
Ionisation energy of hydrogen = 13.6 eV; so, to supply this 13.6 eV energy to the electron of hydrogen atom, minimum 13.6V potential should be applied to it.
Hence, ionisation potential of hydrogen = 13.6 V.
Numerical Examples
Example 1.
Hydrogen atom in its ground state, is excited by means of monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. Given, ionisation energy of hydrogen atom is 13.6 eV.
Solution:
Wavelength of incident radiation,
λ = 975Å = 975 × 10-8 cm
∴ Energy of this photon.
hf = \(\frac{h c}{\lambda}\)
= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{975 \times 10^{-8}}\)erg
= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{975 \times 10^{-8} \times 1.6 \times 10^{-12}}\)eV = 12.74 eV
Ionisation energy of hydrogen atom = 13.6 eV, i.e., the ground state energy of this atom = -13.6 eV. So, the energy of the excited state in which the atom is raised under the influence of incident radiation = -13.6 + 12.74 = -0.86 eV.
The quantum number in the ground state = 1. So, if the quantum number in the excited state be n, then
-0.86 = –\(\frac{13.6}{n^2}\) or, n2 = \(\frac{13.6}{0.86}\) = 16 (approx.)
So, n = 4, i.e., the excited state is the fourth Bohr orbit.
During transition from the fourth Bohr orbit to the ground state, the decrease in energy of the atom may occur in 6 different ways [Fig.]. As a result, 6 lines will be obtained in the spectrum.
Of them, during transition from n = 4 to n = 3, the energy difference is the least and hence in this case, the wavelength of the emitted spectral line will be maximum.
Energy of the third Bohr orbit = –\(\frac{13.6}{3^2}\) = -1.51 eV
∴ Decrease in energy due to transition form n = 4 to n = 3
= -0.86 – (-1.51) = 0.65 eV.
Hence, the relation hf = \(\frac{h c}{\lambda}\) = E4 – E3 gives,
λ = \(\frac{h c}{E_4-E_3}\) = \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{0.65 \times 1.6 \times 10^{-12}}\)
= 19110 × 10-8 cm = 19110Å
The relation between the energy of photon E and wavelength λ after substituting the values of different constants is E (in eV) =
In any calculation, this relation can be used directly.
Success and Failure of the Bohr’s Theory
i) With the help of Bohr’s theory, the spectrum of hydrogen or a hydrogen-like atom can be explained almost accurately.
ii) With the help of Bohr’s theory, the main characteristics of atomic spectrum of alkali metals can be explained.
iii) Bohr’s theory cannot analyse the energy of atoms having more than one electron, nor can it explain the wavelengths of spectral lines of such atoms quantitatively. Still Bohr model is the most suitable model supporting the electronic configuration of any atom.
iv) Actually, every spectral line of the spectrum of hydrogen or other atoms consists of a number of spectral lines; they remain so close to each other that with the help of an ordinary prism-disperser, they cannot be identified separately. This fine structure of every primary spectral line cannot be explained with the help of Bohr’s theory. This can be called the limitation of Bohr’s theory instead of its failure.
v) If a charged particle possesses acceleration, it radiates electromagnetic radiation. This theory of classical physics had been proved by different experiments. But, Bohr’s theory totally fails to explain why no electromagnetic radiation is emitted from the electron having centripetal acceleration while revolving in a Bohr orbit.