Question

Show that the radius of the orbit in hydrogen atoms varies as ${{n}^{2}}$, where n is the principal quantum number of the atom.

Answer 1

Hint: Since, the question talks about the orbit of hydrogen atom so we will use Bohr’s theory, centripetal force and electrostatic forces and after equating required formulas we will get the answer. As this question needs radius so we will not substitute the value of r.

Formula used:
${{F}_{e}}=K\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ where ${{F}_{e}}$ is electrostatic force, ${{q}_{1}},{{q}_{2}}$ are charges, r is separation between charges and K is coulombs’ constant, ${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$ where ${{F}_{c}}$ is called centripetal force, m is mass and v is velocity of an object and r is the radius of the circle.

Complete answer:
Electrostatic force of attraction: These are those types of forces that are exerted between charges. If suppose we have two opposite charges then they will have attractive forces and if they are the same charges then the force that will be between them is going to be repelling in nature. There is also a formula for calculating electrostatic force as ${{F}_{e}}=K\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$.
Bohr’s theory: Bohr’s model is a modified version of Rutherford model in which a positively charged nucleus has a negative charge in its surroundings.
Centripetal force: this is the one of the forces that acts on an object which is moving in a circular direction. This force is always focused on the center of the circular path. The formula of this force is ${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$.
By electrostatic forces of attraction, we have forces between any nucleus and its electron as ${{F}_{e}}=K\dfrac{{{q}^{2}}}{{{r}^{2}}}$.
And on the basis of Bohr’s model we know that a hydrogen atom consists of a positive (+q) as well as a negative charge (-q) which take rounds around the nucleus in only a circular direction. For keeping these charges in its orbit, it is important to have centripetal forces into action. In this case centripetal forces are equal to electrostatic forces. So, we have
$\dfrac{m{{v}^{2}}}{r}=\dfrac{K{{q}^{2}}}{{{r}^{2}}}$
$\Rightarrow m{{v}^{2}}=\dfrac{K{{q}^{2}}}{r}$
$\Rightarrow {{v}^{2}}=\dfrac{K{{q}^{2}}}{mr}$ ….(1)
As we know that the quantization property of Bohr’s model for momentum which is angular is $L=mvr=\dfrac{nh}{2\pi }$ so, we will get ${{m}^{2}}{{v}^{2}}{{r}^{2}}=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}}$ or ${{v}^{2}}=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{r}^{2}}{{m}^{2}}}$ ….(2)
Thus, by (1) and (2) we get
$\dfrac{K{{q}^{2}}}{mr}=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{r}^{2}}{{m}^{2}}}$
$\Rightarrow K{{q}^{2}}=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}rm}$
$\Rightarrow r=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}mK{{q}^{2}}}$
$\Rightarrow r\propto {{n}^{2}}$
Hence, we can clearly see that hydrogen atom orbit’s radius varies as ${{n}^{2}}$, where n is the principal quantum number of the atom means they are directly proportional to each other.

Note:
Take note of the following important points for a solution.
(1) Formula of electrostatic force: ${{F}_{e}}=K\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$.
(2) Formula of centripetal force: ${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$.
(3) The quantization property of Bohr’s model for angular momentum: $L=mvr=\dfrac{nh}{2\pi }$.
(4) Definition of Bohr’s model, electrostatic forces and centripetal forces.