Question
Write the centripetal force formula for an electron moving in a stationary orbit in a hydrogen atom and derive the formula for the radius of the orbit.
✓
Answer
→Form Rutherford's nuclear model of the atom, it can be said that atom is an electrically neutral sphere consisting of a very small, massive and positively charged nucleus at the centre surrounded by the revolving electrons in their respective dynamically stable orbits.
→The electrostatic force of attraction $F _{ e }$ between the revolving electrons and the nucleus provides the required a centripetal force $F _{ c }$ to keep them in their orbits.
→Thus, for a dynamically stable orbit in a hydrogen atom,
$F _{ e }= F _{ c }$
$\begin{aligned}
\therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r^2} & =\frac{m v^2}{r} \\
\therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r} & =m v^2 \\
\therefore \quad r & =\frac{e^2}{4 \pi \varepsilon_0 m v^2} \\
& \text { OR } \\
\therefore \quad v^2 & =\frac{e^2}{4 \pi \varepsilon_0 m r} \\
\therefore & =\sqrt{\frac{e^2}{4 \pi \varepsilon_0 m r}}
\end{aligned}$
→The electrostatic force of attraction $F _{ e }$ between the revolving electrons and the nucleus provides the required a centripetal force $F _{ c }$ to keep them in their orbits.
→Thus, for a dynamically stable orbit in a hydrogen atom,
$F _{ e }= F _{ c }$
$\begin{aligned}
\therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r^2} & =\frac{m v^2}{r} \\
\therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r} & =m v^2 \\
\therefore \quad r & =\frac{e^2}{4 \pi \varepsilon_0 m v^2} \\
& \text { OR } \\
\therefore \quad v^2 & =\frac{e^2}{4 \pi \varepsilon_0 m r} \\
\therefore & =\sqrt{\frac{e^2}{4 \pi \varepsilon_0 m r}}
\end{aligned}$
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