Question

The radius of the second Bohr orbit for hydrogen atom is:
Given:
1. Planck’s constant, $h = 6.6262 \times {10^{ - 34}}\;{\rm{Js}}$
2. Mass of electron, ${m_e} = 9.1091 \times {10^{ - 31}}\;{\rm{kg}}$
3. Charge of electron, $e = 1.60210 \times {10^{ - 19}}\;{\rm{C}}$
4. Permittivity of vacuum, \[{\varepsilon _0} = 8.854185 \times {10^{ - 12}}\;{\rm{k}}{{\rm{g}}^{ - 1}}{{\rm{m}}^{ - 3}}{{\rm{A}}^2}\]
A. \[4.76\;\mathop {\rm{A}}\limits^{\rm{o}} \]
B. \[0.529\;\mathop {\rm{A}}\limits^{\rm{o}} \]
C. \[2.12\;\mathop {\rm{A}}\limits^{\rm{o}} \]
D. \[1.65\;\mathop {\rm{A}}\limits^{\rm{o}} \]

Answer 1

Hint: Bohr’s proposal for quantization of angular momentum of an electron can be used with other known relationships between different variables related to an electron to derive an expression for radius of the second Bohr orbit for hydrogen atom.

Complete step by step solution:
Let’s start with the electrostatic force between the hydrogen nucleus and electron. Hydrogen nucleus has only one proton so it will have a unit charge as $e = 1.60210 \times {10^{ - 19}}\;{\rm{C}}$. Similarly, there is only one electron that also has a unit charge as $e = 1.60210 \times {10^{ - 19}}\;{\rm{C}}$. We can write the expression for the electrostatic force between the hydrogen nucleus and electron as follows:
${\rm{Electrostatic\, force}} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}}$
Here, ${\varepsilon _0}$ is the permittivity of vacuum and $r$ is the distance between the nucleus and the electron that we can also call, radius of a Bohr orbit.
Now, for the stability of the electron, this electrostatic force has to be equal to the centripetal force that results from the circular motion of electron and can be expressed as follows:
${\rm{Centripetal\, force}} = \dfrac{{{m_e}{v^2}}}{r}$
Here, \[{m_e}\] is the mass of an electron and $v$ is the velocity of the electron. Now, let’s equate these two forces as follows:
$\dfrac{{{m_e}{v^2}}}{r} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}}$
As per the quantization of angular momentum, we can write an expression for this as follows:
${m_e}vr = \dfrac{{nh}}{{2\pi }}$
Here, $n$ is the principal quantum number for the orbit and $h$ is the Planck’s constant. We will take the square of both the sides as follows:
${m_e}^2{v^2}{r^2} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}}}$
Let’s rewrite this equation as follows:
$\left( {\dfrac{{{m_e}{v^2}}}{r}} \right){m_e}{r^3} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}}}$
Now, we can substitute the equated expression between the two forces in this expression as follows:
$\left( {\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}}} \right){m_e}{r^3} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}}}$
Let’s simplify this for radius of the Bohr orbit as follows:
\[r = {n^2}\dfrac{{{\varepsilon _0}{h^2}}}{{\pi {m_e}{e^2}}}\]
Finally, we can substitute the given values in this expression to determine the radius of the second Bohr orbit as follows:
\[
r = {n^2}\dfrac{{{\varepsilon _0}{h^2}}}{{\pi {m_e}{e^2}}}\\
 = {\left( 2 \right)^2}\dfrac{{\left( {8.854185 \times {{10}^{ - 12}}\;{\rm{k}}{{\rm{g}}^{ - 1}}{{\rm{m}}^{ - 3}}{{\rm{C}}^2}{{\rm{s}}^2}} \right){{\left( {6.6262 \times {{10}^{ - 34}}\;{\rm{kg}}{{\rm{m}}^2}{{\rm{s}}^{ - 2}}{\rm{s}}} \right)}^2}}}{{\pi \left( {9.1091 \times {{10}^{ - 31}}\;{\rm{kg}}} \right){{\left( {1.60210 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}\\
 = 21.18 \times {10^{ - 11}}\;{\rm{m}}\\
 = 2.12\;\mathop {\rm{A}}\limits^{\rm{o}}
\]

Hence, the correct option is C. \[2.12\;\mathop {\rm{A}}\limits^{\rm{o}} \]

Note:
Radius of \[{n^{{\rm{th}}}}\] Bohr orbit can also be calculated by a simple formula \[{n^2}\left( {52.9\,{\rm{ pm}}} \right)\] as \[52.9\,{\rm{ pm}}\] is the radius of first Bohr orbit.