Question

The expression for the Bohr radius of hydrogen-like species is:
(A) \[\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}Z}}\]
(B) $\dfrac{1}{{{n^2}}}\left[ {\dfrac{{{h^2}}}{{4{\pi ^2}m(Z{e^2}/4\pi {\varepsilon _0})}}} \right]$
(C) ${n^2}\left[ {\dfrac{{4{\pi ^2}m{h^2}}}{{(Z{e^2}/4\pi {\varepsilon _0}}}} \right]$
(D) $\dfrac{1}{{{n^2}}}\left[ {\dfrac{{4{\pi ^2}m{h^2}}}{{(Z{e^2}/4\pi {\varepsilon _0}}}} \right]$

Answer 1

Hint: Neils Bohr was the first Scientist who explained the H-atom structure and its spectrum. He gave many postulates regarding the position of electrons, radius, and energy structure of Hydrogen atoms. The expression for the radius of H-atom can be deduced by equating the centripetal force and the electrostatic force of attraction. We will use the relation between the angular momentum of electrons from one of Bohr’s postulates.

Complete step by step solution:
Let charge on electron be ‘e’
Let the mass of electron be ‘m’
Let the velocity of the electron be ‘v’
Let the radius of the orbit be ‘r’.
Ze is the positive charge on the nucleus of an atom, where Z $ = $ atomic number (For H-atom $Z = 1$).
Since electrons revolve in the orbit with a negative charge on them and the nucleus being in the centre. So here, the centripetal force will provide the necessary electrostatic force of attraction. So now,
$\dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{k(Ze)(e)}}{{{r^2}}}} \right]$
$m{v^2} = \left[ {\dfrac{{k(Z{e^2})}}{{4\pi {\varepsilon _0}r}}} \right].....(i)$
By Bohr’s postulate, the angular momentum of electron is given as:
$mvr = n\dfrac{h}{{2\pi }}.....(ii)$
Where $n = 1,2,3....$ now, Squaring (ii) and dividing by (i) we get,
$r = {n^2}\dfrac{{{h^2}{\varepsilon _0}}}{{\pi mZ{e^2}}}$
For H-atom $Z = 1$, so it will be,
$r = \dfrac{{{n^2}{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}$
The required expression is,
$r = \dfrac{{{n^2}{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}$
Where ‘r’ is the radius of the electron and $n = 1,2,3....$ (Principal quantum number)
${\varepsilon _0} = 8.85 \times {10^{ - 12}}$,$\pi = 3.14$, $h = $Planck’s Constant, $m = $ mass of electron,
Therefore, Option, ‘(D) $\dfrac{1}{{{n^2}}}\left[ {\dfrac{{4{\pi ^2}m{h^2}}}{{(Z{e^2}/4\pi {\varepsilon _0}}}} \right]$’ is the correct answer.

Note: While equating the equations calculations should be done carefully and the correct relation of angular momentum must be used to derive the desired equation in the correct order. If you put all the constant values in the equation for hydrogen first energy level then you will get the value of radius of hydrogen atom will be \[0.529{{ }}{A^0}\]. Neil Bohr also gave expression for different energy levels of the hydrogen spectra.