Question

The radius of the stationary state which is also called Bohr radius is given by the expression $\text{ }{{\text{r}}_{\text{n}}}={{\text{n}}^{2}}{{\text{a}}_{0}}\text{ }$ where the value of $\text{ }{{\text{a}}_{0}}\text{ }$is:
A) $\text{ 52}\text{.9 pm }$
B) $\text{ 5}\text{.29 pm }$
C) $\text{ 529 pm }$
D) $\text{ 0}\text{.529 pm }$

Answer 1

We will solve this question by using the given expression of the radius of the stationary state; it is used for the hydrogen atom and hydrogen-like species. We know Bohr’s theory was valid for hydrogen and similar species. The radius of the orbitals is related to the energy level, a charge of the electron. The $\text{ }{{\text{a}}_{0}}\text{ }$ is proportionality constant.

Complete Solution :
Now, the expression given in the question is used for the hydrogen atom, and also termed as Bohr’s radius i.e. $\text{ }{{\text{r}}_{\text{n}}}={{\text{n}}^{2}}{{\text{a}}_{0}}\text{ }$
Thus, we will write the expression for the radius of hydrogen-like particles in their stationary state, and it can be written as:
$\text{ }{{\text{r}}_{\text{n}}}={{\text{n}}^{2}}\dfrac{{{\text{a}}_{0}}}{\text{Z}}\text{ }$
Here we have n represents the ${{\text{n}}^{\text{th}}}$ Bohr orbit, Z represents the atomic number, and $\text{ }{{\text{a}}_{0}}\text{ }$ represents the radius of the first stationary state, that remains the same in each case.
The above equation can be modified to determine the value $\text{ }{{\text{a}}_{0}}\text{ }$. The equation is,
$\text{ }{{\text{a}}_{0}}=\text{ }\dfrac{{{\text{r}}_{\text{n}}}Z}{{{\text{n}}^{2}}}\text{ }$ (1)

- To determine the value of $\text{ }{{\text{a}}_{0}}\text{ }$, the radius of the first stationary state can be calculated for the hydrogen atom. The hydrogen atom has the atomic number ‘Z’ equal to 1 and the atom has the one energy level $\text{ ( 1s ) }$ . Let's substitute the values in the (1) equation. we have,
$\text{ }{{\text{a}}_{0}}=\text{ }\dfrac{{{\text{r}}_{\text{n}}}Z}{{{\text{n}}^{2}}}\text{ = }\dfrac{{{\text{r}}_{\text{n}}}(1)}{{{(1)}^{2}}}\text{ = }{{\text{r}}_{\text{n}}}\text{ }$
Thus, the hydrogen atom $\text{ }{{\text{a}}_{0}}\text{ }$ is equal to the $\text{ }{{\text{r}}_{\text{n}}}\text{ }$ value.

Thus from the Bohr radius equation, the value of $\text{ }{{\text{a}}_{0}}\text{ }$can be determined as,
$\begin{aligned} & \text{ }{{\text{r}}_{\text{n}}}\text{ }=\text{ }{{\text{a}}_{\text{0}}}\text{ = }\dfrac{{{h}^{2}}4\pi {{\varepsilon }_{0}}}{4{{\pi }^{\text{2}}}{{\text{m}}_{\text{e}}}{{\text{e}}^{\text{2}}}}\text{ } \\\ & \Rightarrow {{\text{a}}_{\text{0}}}=\text{ }\dfrac{\left( 1.112\times {{10}^{-10}}\text{ }{{\text{C}}^{\text{2}}}{{\text{N}}^{-1}}{{\text{m}}^{-2}} \right){{\left( 6.626\times {{10}^{-34}}Js \right)}^{2}}}{4{{\pi }^{\text{2}}}\left( 9.109\times {{10}^{-31}}\text{kg} \right){{\left( 1.602\times {{10}^{-19}}\text{C} \right)}^{2}}}\text{ = }\frac{4.917\times {{10}^{-77}}}{9.219\times {{10}^{-67}}}=0.529\times {{10}^{-10}}\text{m } \\\ \end{aligned}$

- Where h is the planck's constant $\text{ }6.626\times {{10}^{-34}}Js\text{ }$
$\text{ }4\pi {{\varepsilon }_{0}}\text{ }$, is a permittivity factor and its value is equal to $\text{ }1.112\times {{10}^{-10}}\text{ }{{\text{C}}^{\text{2}}}{{\text{N}}^{-1}}{{\text{m}}^{-2}}\text{ }$
$\text{ }{{\text{m}}_{\text{e}}}\text{ }$, is the mass of an electron $\text{ }9.109\times {{10}^{-31}}\text{kg }$
Charge on the electron is ‘e’ is $\text{ }1.602\times {{10}^{-19}}\text{C }$
- So, we can say that the value of $\text{ }{{\text{a}}_{0}}\text{ }$ is $\text{ }0.529\times {{10}^{-10}}\text{m }$ or in the picometer it is equal to $\text{ 52}\text{.9 pm }$.
So, the correct answer is “Option A”.

Note: Note that, the radii of the stationary state mean the first energy level where the value of n is 1. It also means that the energy of the first level will be lowest as it is nearer to the nucleus. The value of radii will be the same for hydrogen-like atoms, moreover in the case of atomic number 1.