Physics > Basic Modern Physics > 11.0 Energy of electron in the $n^{th}$ orbit
Basic Modern Physics
1.0 Photon theory of light
2.0 Characteristics of photon
3.0 Wave Particle Duality
4.0 Emission of electrons
5.0 Photoelectric Effect
5.1 Laws of Photoelectric emission
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
6.0 Radiation Pressure And Force
7.0 Photon Density
8.0 Force exerted by a light beam on a surface
9.0 Early Atomic Structures
10.0 Bohr Model of The Hydrogen Atom
10.1 Radius of Orbit
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
11.0 Energy of electron in the $n^{th}$ orbit
12.0 Basic Definitions
13.0 Atomic Excitation
11.1 Hypothetical Atomic Energy Level
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
If given an atom where electron is revolving under the influence of a conservative potential energy field $U = {\text{f}}\left( r \right)$ instead of the coulomb force field, we can use the following Bohr's postulates to calculate the expressions for the radius, velocity and energy of the hypothetical atom.
$$\begin{equation} \begin{aligned} 1.\ \ F\left( { = \left| { - \frac{{\partial U}}{{\partial r}}} \right|} \right) = \frac{{mv_n^2}}{{{r_n}}} \\ 2.\ \ m{v_n}{r_n} = \frac{{nh}}{{2\pi }} \\\end{aligned} \end{equation} $$
Question 13. Suppose the potential energy between the electron and proton at separation $r$ is given by $U = k\log r$, where $k$ is a constant. For such a hypothetical hydrogen atom, calculate the radius of the $n$th Bohr's orbit and its energy levels.
Solution: For a conservative force field is $$ F = - \frac{{dU}}{{dr}} = - \frac{k}{r} $$
This force $F = - k/r$ provides the centripetal force for the circular motion of electron.
$$\begin{equation} \begin{aligned} {\text{So, }}\frac{{m{v^2}}}{r} = \frac{k}{r}\quad \Rightarrow \quad {m^2}{v^2} = mk\quad \Rightarrow \quad mv = \sqrt {mk} ...(i) \\ {\text{Applying Bohr's quantization rule, }}mvr = \frac{{nh}}{{2\pi }}...(ii) \\ {\text{From equation (i) and (ii), we get }}r = \frac{{nh}}{{2\pi \sqrt {mk} }} \\ {\text{From Eq}}{\text{.(i), KE of electron }} = \frac{1}{2}m{v^2} = \frac{1}{2}k \\ {\text{Total energy of electron }} = KE + PE = \frac{1}{2}k + k\log r \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{k}{2}\left[ {1 + \log \frac{{{n^2}{h^2}}}{{4{\pi ^2}mk}}} \right] \\\end{aligned} \end{equation} $$
