Derivation of Bohr's radii and energies from electrostatics + quantization

1.2.8 · Chemistry › Atomic Structure (Classical)

WHAT are we deriving?

• Electron ki circular orbit ki allowed radii $r_n$.
• Atom ki allowed total energies $E_n$.

Hum yeh ek hydrogen-like atom ke liye karte hain: ek electron, nuclear charge $+Ze$ (H ke liye $Z=1$, He${}^{+}$ ke liye $Z=2$, ...).

HOW: the two ingredients

WHY these two? Ingredient 1 pure classical physics hai (jo electron ko circling rakhta hai). Ingredient 2 woh naya quantum rule hai jo Bohr ne bolt on kiya — taaki electron nucleus mein spiral karke na gire aur sharp spectral lines explain ho sakein.

Derivation of the radius $r_n$

Step 1 — Simplify Coulomb = centripetal. Dono sides ko $r$ se multiply karo: \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r} = m v^2 \tag{A} Why this step? Hume $r$ aur $v$ ke beech ek aisa relation chahiye jo quantization ke saath combine ho sake.

Step 2 — Get $v$ from quantization. $mvr=\frac{nh}{2\pi }$ se: v = \frac{nh}{2\pi m r} \tag{B} Why this step? Quantization rule woh ek maatra extra information hai jo $r$ ki value pin down karega.

Step 3 — Substitute (B) into (A). $\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}=m{\left(\frac{nh}{2\pi mr}\right)}^2=\frac{n^2{h}^2}{4{\pi }^{2}m{r}^2}$ Why this step? $v$ ko khatam karo taaki sirf $r$ bache.

Step 4 — Solve for $r$. Ek $r$ cancel karo aur rearrange karo: $\frac{Z{e}^2}{4\pi {\epsilon }_0}=\frac{n^2{h}^2}{4{\pi }^{2}mr}\Rightarrow r=\frac{{\epsilon }_0{n}^2{h}^2}{\pi mZ{e}^2}$

Read it off: radius $==n^2==$ ke saath badhti hai aur $Z$ badhne par ghatti hai (bada nuclear charge electron ko aur kareeb kheench leta hai).

Derivation of the energy $E_n$

Step 1 — Write total energy. $E=KE+PE$.

• Kinetic: $KE=\frac{1}{2}m{v}^2$.
• Potential (Coulomb, attractive isliye negative): $PE=-\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}$.

Step 2 — Use (A) to relate KE and PE. (A) se, $m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}$, toh $KE=\frac{1}{2}m{v}^2=\frac{1}{2}\cdot \frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}=-\frac{1}{2}PE$ Why this step? Yeh $1/r$ force ke liye virial result hai — yeh humein sab kuch $PE$ ke terms mein likhne deta hai.

Step 3 — Total energy. $E=KE+PE=-\frac{1}{2}PE+PE=\frac{1}{2}PE=-\frac{1}{2}\cdot \frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}$ $E=-\frac{Z{e}^2}{8\pi {\epsilon }_{0}r}$ Why this step? Energy potential energy ki half hai — aur yeh negative hai, matlab electron bound hai.

Step 4 — Plug in $r_n$. $r_n=\frac{{\epsilon }_0{n}^2{h}^2}{\pi mZ{e}^2}$ substitute karo: $E_n=-\frac{Z{e}^2}{8\pi {\epsilon }_0}\cdot \frac{\pi mZ{e}^2}{{\epsilon }_0{n}^2{h}^2}=-\frac{m{Z}^2{e}^4}{8{\epsilon }_0^2{n}^2{h}^2}$

Read it off: energy negative hai (bound), $==Z^2/n^2==$ ke saath scale karti hai, aur $n$ badhne par $0$ ke kareeb aati jaati hai (kam bound hoti hai).

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Connections

• Bohr Model Postulates
• Coulomb's Law
• Centripetal Force and Circular Motion
• Quantization of Angular Momentum
• Hydrogen Spectrum and Rydberg Formula
• Ionization Energy
• Virial Theorem
• de Broglie Wavelength and Standing Waves (quantization ke liye physical justification)

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