Derivation of the Energy of the Bohr model of a hydrogen atom | Quantum Science Philippines
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Derivation of the Energy of the Bohr model of a hydrogen atom

 Problem: Derive E_n= -\frac{1}{n^2} E_I

 

Solution

We start with the three equations below,

\textit{Equation (1) : } E = \frac {1}{2} \mu {v}^2 - \frac{{e}^2}{r}

 

\textit{Equation (2) : } \frac {\mu {v}^2}{r}= \frac{{e}^2}{{r}^2}

 

\textit{Equation (3): } \mu vr = n\hbar

 

Equation 1 is an expression of the fact that the total energy E of an electron is the sum of its kinetic energy \frac{\mu v^2}{2} and its potential energy -\frac{e^2}{r} . Equation 2 is the fundamental equation of Newtonian dynamics, \frac{{e}^2}{r} is the Coulomb force exerted on the electron and \frac{v^2}{r} is the acceleration of its uniform circular motion. Lastly, equation 3 expresses the quantization condition introduced by Bhor to explain the existence of discrete energy levels.

From equation 2,

\begin{array}{ccc} \frac {\mu {v}^2}{r}= \frac{{e}^2}{{r}^2} \\ \mu {v}^2= \frac{{e}^2}{r} \end{array}

Substituting this to equation 1,

 

\begin{array}{ccc} E &=& \frac {1}{2} \mu {v}^2 - \frac{e^2}{r}\\ &=& \frac{1}{2} \frac{e^2}{r}-\frac{e^2}{r}\\ &=& -\frac{1}{2} \frac{e^2}{r}\end{array}

 

Now we solve for r from equation 3 then substitute it to the E above.

\begin{array}{ccc} \mu vr = n \hbar \\ r= \frac{n\hbar}{\mu v} \end{array}

 

E= -\frac{1}{2} e^2 \bigg( \frac{\mu v}{n\hbar} \bigg) = -\frac{1}{2} \frac{\mu v{e}^2}{n\hbar}

 

From equation 2, we solve for v,

 

\begin{array}{ccc} \frac{\mu v^2}{r} =\frac{e^2}{r^2} \\ v= \frac{e^2}{\mu vr} \end{array}

Finally we substitute v to E,

 

\begin{array}{ccc} E &=& -\frac{1}{2} \frac{\mu v{e}^2}{n\hbar} \\ &=& -\frac{1}{2} \frac{\mu {e}^2}{n\hbar}\bigg(\frac{e^2}{\mu vr} \bigg) \end{array}

 

recall r= \frac{n\hbar}{\mu v}

\begin{array}{ccc}E&=& -\frac{1}{2} \frac{e^4}{\mu n \hbar} \bigg( \frac{\mu v}{n \hbar} \bigg) \\ &=& -\frac{1}{n^2} \frac{\mu {e}^{4}}{n{\hbar}^{2}} \end{array}

Where we have the Ionization Energy E_I=\frac{\mu {e}^{4}}{n{\hbar}^{2}}

 

Thus,

 

E_n= -\frac{1}{n^2} E_I

 

 

 

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