Question

Bohr's theory does not apply to: THIS QUESTION HAS MULTIPLE CORRECT OPTIONS
A.He
B.\[L{i^{ + 2}}\]
C.\[H{e^{ + 2}}\]
D.the H-atom

Answer 1

Hint: Bohr was able to derive an equation that matched the relationship obtained from the analysis of the spectrum of the atom. According to his model, the frequency of the light emitted by an atom when the electron falls from a high energy orbit into a lower energy orbit.

Complete step by step answer:
Planck's equation states that the energy of a photon is proportional to its frequency i.e. \[\Delta E = h\nu \]. From Bohr’s model, substituting the relationship between the frequency, wavelength and the speed of light into this equation tells us that the energy of a photon is inversely proportional to its wavelength and the inverse of the wavelength of electromagnetic radiation is directly proportional to the energy of this radiation. So, we can say that frequency is directly proportional to the energy of this radiation.
We know the formula of change in energy and this can be used to determine the frequency and wavelength of the light emitted by a hydrogen atom when the electron falls from a high energy orbit into a lower energy orbit as
 \[\Delta E = h\nu = \dfrac{{2{\pi ^2}m{Z^2}{e^4}{k^2}}}{{{h^2}}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\] if we see that electron falls from \[{n_2}\] level to \[{n_1}\] level. The theory of Bohr is only applicable for 1 electron species like hydrogen, \[L{i^{ + 2}}\], \[H{e^ + }\].

According to the given options, the correct answer is, B, D.

Note: The Rydberg Constant equation was shown by Niels Bohr. In the formulae \[{R_H} = \dfrac{{m{k^2}{e^4}}}{{4\pi {h^3}c}}\] for the wavenumbers of lines in atomic spectra Rydberg constant appears. It is a function of the rest mass and charge of the electron, the speed of light, and Planck’s constant. Its value is \[1.09677576 \times {10^7}{m^{ - 1}}\].