Question

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as \[F = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right)\], where \[\beta \] is a constant. For this atom, the radius of the nth orbit in terms of the Bohr radius \[\left( {{a_0} = \dfrac{{{\varepsilon _0}{h^2}}}{{m\pi {e^2}}}} \right)\] is:
A. \[{r_n} = {a_0}n - \beta \]
B. \[{r_n} = {a_0}{n^2} + \beta \]
C. \[{r_n} = {a_0}{n^2} - \beta \]
D. \[{r_n} = {a_0}n + \beta \]

Answer 1

Hint:To solve this question we need to use the postulate of Bohr’s model. According to Bohr's model, electrons are revolving around the nucleus in a circular orbit in an atom. After comparing the postulates of Bohr’s model with the given radius of the atom we will get the resultant relation.

Formula used:
The second postulate of Bohr’s model is given as,
\[mvr = \dfrac{{nh}}{{2\pi }}\]
Where m is the mass of an electron, v is the velocity of an electron, r is the radius of nth orbit, n is an integer and h is the Planck’s constant.
In a circular motion, the centripetal force is given as,
\[F = \dfrac{{m{v^2}}}{r}\]
Where m is the mass of the revolving element, v is the velocity and r is the radius of circular orbit.

Complete step by step solution:
Given that the force between the nucleus and the electron,
\[F = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right)\]
As the electron is moved in a circular orbit around the nucleus with a centripetal force. So, this force is equal to
\[\dfrac{{m{v^2}}}{r} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right) \\ \]
\[\Rightarrow {v^2} = \dfrac{{r{e^2}}}{{4m\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right) \\ \]
As according to the second postulate of Bohr’s model,
\[mvr = \dfrac{{nh}}{{2\pi }} \\ \]
\[\Rightarrow v = \dfrac{{nh}}{{2\pi mr}}\]

Using this value in above equation,
\[{\left( {\dfrac{{nh}}{{2\pi }}} \right)^2} = \dfrac{{r{e^2}}}{{4m\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right) \\ \]
By solving this we have
\[\dfrac{{{\varepsilon _0}{n^2}{h^2}}}{{\pi m{r^3}{e^2}}} = \left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right) \\ \]
Now using the radius of the nth orbit in terms of the Bohr radius which is given as,
\[\left( {{a_0} = \dfrac{{{\varepsilon _0}{h^2}}}{{m\pi {e^2}}}} \right) \\ \]
\[\Rightarrow \dfrac{{{a_0}{n^2}}}{{{r^3}}} = \left( {\dfrac{1}{{{r^2}}} + \dfrac{\beta }{{{r^3}}}} \right) \\ \]
By solving this we have
\[\therefore r = {a_0}{n^2} - \beta \]

Hence option C is the correct answer.

Note: Bohr’s model of the atom was given by Neil Bohr in 1915 by modifying Rutherford’s model of an atom. The Bohr model of an atom came into existence with some modification of Rutherford’s model of an atom. Bohr’s theory modified the atomic structure of the model by explaining that electrons will move in fixed orbitals or shells and each of the orbitals or shells has its fixed energy.