Question

Write down Bohr's postulates for hydrogen atom. Prove that orbital radius (r) of hydrogen atom is directly proportional to square of the principal quantum number (n).

Hint:

The derivation for Bohr's radius and energy levels relies on two fundamental equations: the force balance equation (\(F_{centripetal} = F_{electric}\)) and the angular momentum quantization condition (\(mvr = nh/2\pi\)). By combining these two, you can derive expressions for \(r\), \(v\), and \(E\) in terms of \(n\).

Answer 1

1. Bohr's Postulates for the Hydrogen Atom:
\begin{enumerate} \item Postulate of Stationary Orbits: An electron in an atom can revolve in certain stable, circular orbits called stationary orbits without emitting any radiation. Each orbit has a definite energy associated with it. \item Postulate of Quantization of Angular Momentum: The angular momentum (\(L\)) of an electron in a stationary orbit is an integral multiple of \(h/2\pi\), where \(h\) is Planck's constant. \[ L = m_e v r = n \frac{h}{2\pi} \] where \(n = 1, 2, 3, \ldots\) is the principal quantum number. \item Postulate of Energy Transition: An electron can make a transition from a higher energy stationary orbit to a lower energy one. When it does so, a photon is emitted whose energy is equal to the energy difference between the initial (\(E_i\)) and final (\(E_f\)) orbits. \[ E_{photon} = h\nu = E_i - E_f \] \end{enumerate} 2. Proof that \(r \propto n^2\):
Consider an electron of mass \(m_e\) and charge \(-e\) revolving around a nucleus of charge \(+e\) (for hydrogen, Z=1) in a circular orbit of radius \(r\) with a velocity \(v\).
\begin{itemize} \item Force Balance: The electrostatic force of attraction between the nucleus and the electron provides the necessary centripetal force for the circular motion. \[ F_{centripetal} = F_{electrostatic} \] \[ \frac{m_e v^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \cdots (1) \] \item Bohr's Quantization Condition: From the second postulate: \[ m_e v r = \frac{nh}{2\pi} \] From this, we can express the velocity \(v\) as: \[ v = \frac{nh}{2\pi m_e r} \cdots (2) \] \item Substitution: Substitute the expression for \(v\) from equation (2) into equation (1): \[ \frac{m_e}{r} \left( \frac{nh}{2\pi m_e r} \right)^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \] \[ \frac{m_e}{r} \frac{n^2 h^2}{4\pi^2 m_e^2 r^2} = \frac{e^2}{4\pi\epsilon_0 r^2} \] \item Simplification and Solving for \(r\): \[ \frac{n^2 h^2}{4\pi^2 m_e r^3} = \frac{e^2}{4\pi\epsilon_0 r^2} \] Cancel \(4\pi\) and \(r^2\) from both denominators: \[ \frac{n^2 h^2}{\pi m_e r} = \frac{e^2}{\epsilon_0} \] Rearranging the terms to solve for the radius \(r\): \[ r = \frac{n^2 h^2 \epsilon_0}{\pi m_e e^2} \] \item Conclusion: In this expression, \(h, \epsilon_0, \pi, m_e,\) and \(e\) are all fundamental constants. Therefore, we can write the radius as: \[ r = (\text{constant}) \times n^2 \] This proves that the radius of the orbital is directly proportional to the square of the principal quantum number (\(r \propto n^2\)). \end{itemize}