NCERT Solutions for Chapter: Atoms, Exercise 1: MCQ

Taking the Bohr radius as $r_0 = 53 \mathrm{pm},$ the radius of $L{i}^{++}$ ion in its ground state, on the basis of Bohr’s model, will be about

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is $B=-\frac{m{e}^4}{8n^2{{\epsilon }_0}^2{h}^2}$. ($m=$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

$B=-\frac{M{e}^4}{8n^2{{\epsilon }_0}^2{h}^2}(M=\mathrm{proton} \mathrm{mass})$ 

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:

According to the simple Bohr model, for the ground state, the electron in the H-atom has an angular momentum $\sout{h}$. 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.

$O_2$ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

Two $H$ atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

A set of atoms in an excited state decays.