To calculate the size of a hydrogen ion (H^-) | vedclass

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(B) Let the velocity of the electron be $v$.The angular momentum of each electron is given as $mvr = \hbar$.The net electrostatic force acting on one

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$\frac{4}{3} a_B$

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(B) Let the velocity of the electron be $v$.The angular momentum of each electron is given as $mvr = \hbar$.The net electrostatic force acting on one electron is the attraction from the nucleus minus the repulsion from the other electron. Since the electrons are on opposite sides,the distance between them is $2r$.$F_e = \frac{e^2}{4\pi\varepsilon_0 r^2} - \frac{e^2}{4\pi\varepsilon_0(2r)^2} = \frac{e^2}{4\pi\varepsilon_0 r^2} - \frac{e^2}{16\pi\varepsilon_0 r^2} = \frac{3}{4} \frac{e^2}{4\pi\varepsilon_0 r^2}$.The centripetal force required for circular motion is $F_c = \frac{mv^2}{r} = \frac{(mvr)^2}{mr^3} = \frac{\hbar^2}{mr^3}$.Equating the forces,$F_c = F_e$:$\frac{\hbar^2}{mr^3} = \frac{3}{4} \frac{e^2}{4\pi\varepsilon_0 r^2}$.Solving for $r$:$r = \frac{4}{3} \left( \frac{4\pi\varepsilon_0\hbar^2}{me^2} \right) = \frac{4}{3} a_B$.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

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