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,

$\mathrm{B}=-\frac{{\mathrm{me}}^4}{8{\mathrm{n}}^2{\mathrm{\epsilon }}_0^2{\mathrm{h}}^2}$ (\(\mathrm{m}=\) 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,

$\mathrm{B}=-\frac{{\mathrm{Me}}^4}{8{\mathrm{n}}^2{\mathrm{\epsilon }}_0^2{\mathrm{h}}^2}$ (\(\mathrm{M}=\) 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 = $\frac{\mathrm{h}}{2\mathrm{\pi }}$, 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,
 

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

1. (a), (b)
2. (c), (d)
3. (b), (c)
4. (a), (d)

Let $E_n=\frac{-1}{8{\mathrm{\epsilon }}_0^2}\frac{m{e}^4}{n^2{h}^2}$ be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $\frac{(E_2-E_1)}{h}$ falls on it,

(a) it will not be absorbed at all
(b) some of the atoms will move to the first excited state
(c) all atoms will be excited to the n = 2 state
(d) no atoms will make a transition to the n = 3 state

1. (b, d)

2. (a, d)

3. (b, c, d)

4. (c, d)

The simple Bohr model is not applicable to He⁴ atom because

(a) He⁴ is an inert gas
(b) He⁴ has neutrons in the nucleus
(c) He⁴ has one more electron
(d) electrons are not subject to central forces

1. (a, c)

2. (a, c, d)

3. (b, d)

4. (c, d)