Question

: If the radius of secondary stationary orbit (in Bohr’s atom) is ${\text{R}}$. Then the radius of third orbit will be:
A.$\dfrac{{\text{R}}}{3}$
B.$9{\text{R}}$
C.$\dfrac{{\text{R}}}{9}$
D.$2.25{\text{R}}$

Answer 1

For this we need to know the formula for calculation of radius of hydrogen and hydrogen like species. For secondary orbit the value of $n$ will be 2 and for the third stationary state value of $n$ will be 3.
Formula used:
${{\text{r}}_n} = 0.529\dfrac{{{n^2}}}{Z}$
Where, ${{\text{r}}_n}$ is radius of stationary states of ${n^{th}}$ orbit, $n$ is the number of stationary state and $Z$ is atomic number of element.

Complete step by step answer:
According to Bohr, atoms consist of various stationary states whose energy is fixed or quantized. The radius of these orbits varies according to the number of the placement. The formula for calculation of radius was given by Bohr.
Putting the value of $n = 2$ in the above equation, we will get the value of the second stationary orbit. It will be:

$${{\text{r}}_2} = 0.529\dfrac{{{2^2}}}{Z} \\\ = \dfrac{{2.116}}{Z} \\\$$

Since it is given to us that ${{\text{r}}_2}$ is ${\text{R}}$, substituting this in above equation:
${\text{R}} = \dfrac{{2.116}}{Z}$ ……equation 1
Now, putting the value of $n = 3$ in the same equation, we will get

{{\text{r}}_3} = 0.529\dfrac{{{3^2}}}{Z} \\\ = \dfrac{{4.761}}{Z} \\\ $$ ……equation 2 Dividing the equation 2 by equation 1, we will get

\dfrac{{{{\text{r}}_3}}}{R} = \dfrac{{\dfrac{{4.761}}{Z}}}{{\dfrac{{2.116}}{Z}}} \\
= \dfrac{{4.761}}{{2.116}} \\
= 2.25 \\

Hence, we will get: $$\dfrac{{{{\text{r}}_3}}}{R} = 2.25$$ Rearranging we will get, $${r_3} = 2.25R$$ **Hence, the correct option is D.** **Note:** There is a term named as Bohr radius. It is a constant quantity. It is represented as $${{\text{a}}_{\text{0}}}$$. Its value is $$5.291{\text{ }} \times {\text{ }}{10^{ - 11}}$$ m. Though the Bohr model is no longer used, the quantum model is used because it gives most practical and verified results. But Bohr radius is very much used because of its vast uses in various derivations.