Question

The radius of the first Bohr orbit is ${a_0}$​ . The \[{n^{th}}\] orbit has a radius:
A) $n{a_0}$
B) $\dfrac{{{a_0}}}{n}$
C) ${n^2}{a_0}$
D) $\dfrac{{{a_0}}}{{{n^2}}}$

Answer 1

Hint: The radius of Bohr’s first orbit is given, we know Bohr gave the model for hydrogen atom. We can use the formula given by him for the radius of \[{n^{th}}\] orbit. From there, we can find the required value of radius.
The known value of first orbit’s radius of hydrogen atom is $0.529\mathop A\limits^\circ $

Complete step by step answer:
According to Bohr’s model of hydrogen atom, the radius of the orbit for \[{n^{th}}\] orbit is given as:
${r_n} = \dfrac{{{\varepsilon _0}{n^2}{h^2}}}{{\pi mZ{e^2}}}$ where,
n is the principal quantum number of the orbit, h is Planck’s constant, e is charged on an electron, r is the radius and Z is atomic number.
All the other quantities on R.H.S in the above equation are constant except n and Z. so, radius will be directly proportional to these (in the form as in the equation)
$ \Rightarrow {r_n} \propto \dfrac{{{n^2}}}{Z}$
The value of $\dfrac{{{\varepsilon _0}{h^2}}}{{\pi m{e^2}}}$ is $0.529\mathop A\limits^\circ $
$ \Rightarrow {r_n} = 0.529 \times \dfrac{{{n^2}}}{Z}$
The radius of first orbit of hydrogen atom has known value of $0.529\mathop A\limits^\circ $ and the question, it is given to be ${a_0}$, so the equation becomes:
$ \Rightarrow {r_n} = {a_0} \times \dfrac{{{n^2}}}{Z}$
The atomic number of hydrogen atoms is 1 so the value of Z also becomes 1. The value of radius in \[{n^{th}}\] orbit will be given by:
 $ \Rightarrow {r_n} = {n^2}{a_0}\left( {\because Z = 1} \right)$

So, the correct answer is “Option C”.

Note:
The formula of Bohr’s radius used here is applicable to isoelectronic to hydrogen i.e. having the same number of electrons as that in hydrogen atom.
We generally denote Bohr’s radius by ${a_0}$ only or sometimes by ${r_{bohr}}$.
$\mathop A\limits^\circ $ used in the question is the units called angstrom and the radius of orbits in atoms are generally measured in this unit only because it is a very small quantity compared to metres. $\left( {1\mathop A\limits^\circ = {{10}^{ - 11}}m} \right)$