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Consider a hypothetical atom having atomic number \[z=2\]. It has two electrons each having mass \[m\]. Assume that both electrons always lie diametrically opposite and the nucleus of the atom contains two protons. Assume the Bohr model is applicable. If the radius of the first orbit of this atom is \[\left( \dfrac{15}{n} \right)pm\] (up to one decimal place in pm). Find the value of \[n\text{ }\left[ \text{given}\;\dfrac{{{\varepsilon }_{0}}{{h}^{2}}}{m{{e}^{2}}}=1.65A{}^\circ \right]\]

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Last updated date: 17th Apr 2024
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Answer
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Hint: To find the radius of the orbit, we can use the property that any rotating particle, here an electron, experiences a centrifugal force. This centrifugal force must be balanced for the electron to maintain its circular motion. The balancing force will be compensated by the electrostatic interaction between the nucleus and the electron. We might also need to use the angular momentum concept for a rotating object.

Formula Used:
 \[F=k\times \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\], \[mvr=n\times \dfrac{h}{2\pi }\], \[{{F}_{c}}=\dfrac{m{{v}^{2}}}{r}\]

Complete step by step solution:
Let the velocity of the electron in the first orbit be \[v\text{ }m/s\]
Angular momentum of the electron can be given as the product of the mass of the electron, its velocity and the radius of the orbit and is given as
\[mvr=\dfrac{h}{2\pi }\]
Now we will consider the electrostatic interaction of each electron; the electron will be attracted by the protons in the nucleus and repelled by the other electron present in the orbit
The net electrostatic interaction experienced by the electron due to the nucleus will be given as
\[{{F}_{e}}=k\times \dfrac{2{{e}^{2}}}{{{r}^{2}}}-k\times \dfrac{{{e}^{2}}}{{{(2r)}^{2}}}\] where \[k\] is the constant of proportionality and has the value \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}\], \[e\] is the elementary charge present on a proton or an electron, \[k\times \dfrac{2{{e}^{2}}}{{{r}^{2}}}\] is the force of attraction between the nucleus and the electron and \[k\times \dfrac{{{e}^{2}}}{{{(2r)}^{2}}}\] is the repulsion offered from the other revolving electron
Note that difference is taken because the forces are acting in opposite directions.
Simplifying the above equation, we get
\[\begin{align}
  & {{F}_{e}}=k\times \dfrac{{{e}^{2}}}{{{r}^{2}}}\left[ 2-\dfrac{1}{4} \right] \\
 & \Rightarrow {{F}_{e}}=k\times \dfrac{7{{e}^{2}}}{4{{r}^{2}}} \\
\end{align}\]
The centrifugal force experienced by the revolving electron can be given as
\[{{F}_{c}}=\dfrac{m{{v}^{2}}}{r}\] where the meanings of the symbols have been discussed above
Multiplying both numerator and denominator by \[m{{r}^{2}}\], we get
\[{{F}_{c}}=\dfrac{{{(mvr)}^{2}}}{r\times m{{r}^{2}}}\]
We know the value of the numerator as angular momentum and we wrote an expression for it above; substituting the value of the angular momentum, we get
\[{{F}_{c}}=\dfrac{{{\left( \dfrac{h}{2\pi } \right)}^{2}}}{r\times m{{r}^{2}}}\]
For the electron to remain in orbit, the electrostatic interaction must be balanced by the centrifugal force due to revolution
Mathematically, we can express this as
\[{{F}_{c}}={{F}_{e}}\]
Substituting the values of the electrostatic force of attraction and the centrifugal force, we get
\[\dfrac{{{\left( \dfrac{h}{2\pi } \right)}^{2}}}{r\times m{{r}^{2}}}=k\times \dfrac{7{{e}^{2}}}{4{{r}^{2}}}\]
Substituting the value of \[k\], we get
\[\dfrac{{{\left( \dfrac{h}{2\pi } \right)}^{2}}}{r\times m{{r}^{2}}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{7{{e}^{2}}}{4{{r}^{2}}}\]
Further simplifying and rearranging, we get
\[\begin{align}
  & \dfrac{{{h}^{2}}}{4{{\pi }^{2}}mr}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{7{{e}^{2}}}{4} \\
 & \Rightarrow \dfrac{{{h}^{2}}}{\pi mr}=\dfrac{1}{{{\varepsilon }_{0}}}\times \dfrac{7{{e}^{2}}}{4} \\
 & \Rightarrow \dfrac{7r}{4}=\dfrac{{{\varepsilon }_{0}}{{h}^{2}}}{m{{e}^{2}}}\times \dfrac{1}{\pi } \\
 & \Rightarrow r=\dfrac{4}{7\pi }\times \dfrac{{{\varepsilon }_{0}}{{h}^{2}}}{m{{e}^{2}}} \\
\end{align}\]

We have been given the value of \[\dfrac{{{\varepsilon }_{0}}{{h}^{2}}}{m{{e}^{2}}}\] as \[1.65A{}^\circ \] or \[1.65\times {{10}^{-10}}m\left[ \because 1A{}^\circ ={{10}^{-10}}m \right]\]
Substituting this value in the above expression, we get
\[\begin{align}
  & r=\dfrac{4}{7\pi }\times 1.65\times {{10}^{-10}} \\
 & \Rightarrow r=0.3001\times {{10}^{-10}}m \\
 & \Rightarrow r=30\times {{10}^{-12}}m=30pm\left[ \because 1m={{10}^{-12}}m \right] \\
\end{align}\]
We have been given the value of the radius of the orbit as \[\left( \dfrac{15}{n} \right)pm\]
Comparing the given value with the obtained value, we get
\[\begin{align}
  & \left( \dfrac{15}{n} \right)pm=30pm \\
 & \Rightarrow \left( \dfrac{15}{n} \right)=30 \\
 & \Rightarrow n=\dfrac{15}{30}=0.5 \\
\end{align}\]
This is the required value of n.

Note: The angular momentum of the electron is always given as an integral multiple of \[\dfrac{h}{2\pi }\] ; but since we are considered with the first orbit only, we took the multiplication factor as one. You should be aware of unit conversion, for example, we had to convert Armstrong into Picometer so we needed the conversion factor. The value of some terms is given in the equation; we must always try to express our answer in terms of that term.