Explore topic-wise InterviewSolutions in .

A source of hydrogen spectrum has billions of hydrogen atoms. Each hydrogen atom has many stationary states. All possible transitions can occur from any higher level to any lower level. This gives rise to a large number of spectral lines.

Limitations of Bohr Model:

Bohr model has many limitations. Some are given here:

1. The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to more two electron atoms such as helium. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. Difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons.

2. It does not explain why only circular orbits should be chosen when elliptical orbits are also possible.

3. In the spectrum of hydrogen, certain spectral lines are not single lines but a group of closed lines with slightly different frequencies. Bohr’s theory could not explain these fine structures of the hydrogen spectrum.

4. Bohr’s theory does not tell anything about the relative intensities of the various spectral lines. Bohr’s theory predicts only the frequencies of these lines.

5. It does not explain the further splitting of spectral lines in a magnetic field (Zeeman effect) or in an electric field (Stark effect).

6. The formulation of Bohr model involves electrical force between positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms.

L = \(n \frac{h}{2 \pi}\)
where, = L = mv_nr_n
mv_nr_n = \(\frac{n h}{2 \pi}\)

de-Broglie matter wave hypothesis.

We get only Lyman series in absorption spectrum of hydrogen atom. Because in normal state all atoms are in lower state (n = 1).

In ultraviolet region.

(c) angular momentum of electron

The answer is (a) decreases

r_n = n²r₁
r₄ = (4)² (0.5) Å
r₄ =(16 × 0.5) Å
r₄ = 8.0 Å

r_n = \(\frac{\epsilon_{0} h^{2}}{\pi m Z e^{2}} n^{2}\)
r_n = n²r₁
r₁ : r₂ : r₃ = (1)² : (2)² : (3)² ……….
= 1 : 4 : 9 ……….

Potential energy U_n = \(-\frac{K Z e^{2}}{r_{n}}\)
Total energy E_n = \(-\frac{1}{2} \frac{K Z e^{2}}{r_{n}}\)
E_n = \(\frac{1}{2}\) (U_n)
U_n = 2E_n
Total energy in first orbit E_n = -13.6 eV
U_n = 2(-13.6)eV
U_n = -27.2 eV

Rutherford a-ray scattering experiment.