Taking the bohr radius as \(a_0=53\) pm, the radius of Li⁺⁺ ion in its ground state on the basis of bohr's model will be about:
1. \(153\) pm
2. \(27\) pm
3. \(18\) pm
4. \(13\) pm

The binding energy of a H-atom, considering an electron moving around a fixed nucleus (proton), is,

\(B = - \dfrac{me^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (\(\mathrm{m}=\) electron mass)
If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be,

\(B = - \dfrac{me^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (\(\mathrm{M}=\) proton mass)
This last expression is not correct, because,

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because:

For the ground state, the electron in the H-atom has an angular momentum \(\dfrac h{2\pi}\), according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

A set of atoms in an excited state decays

An ionised \(H\)-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state:

The Bohr model for the spectra of a \(H\)-atom:

The Balmer series for the H-atom can be observed:
 

1. (b, c)

2. (a, c)

3. (b, d)

4. (c, d)

Let \(E_{n} = \dfrac{- 1}{8 \varepsilon_{0}^{2}} \dfrac{m e^{4}}{n^{2} h^{2}}\) be the energy of the \(n^\text{th}\) level of H-atom. If all the H-atoms are in the ground state and radiation of frequency \(\dfrac{\left(\right. E_{2} - E_{1} \left.\right)}{h}\) falls on it, then:

The simple Bohr model is not applicable to \(\text{He}^4\) atom because:

 
Choose the correct option: