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(D) According to Bohr's model, the electrostatic force provides the necessary centripetal force for the electron: $F = \frac{1}{4\pi\epsilon_0} \frac{

Vedclass · Accepted Answer

$8$

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(D) According to Bohr's model, the electrostatic force provides the necessary centripetal force for the electron: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2} = ma$. Thus, acceleration $a = \frac{Ze^2}{4\pi\epsilon_0 m r^2}$. From Bohr's theory, the radius $r$ is given by $r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$, which implies $r \propto \frac{1}{Z}$. Substituting $r \propto \frac{1}{Z}$ into the acceleration formula: $a \propto Z \cdot (\frac{1}{r^2}) \propto Z \cdot Z^2 = Z^3$. For Hydrogen $(H)$, $Z_H = 1$. For singly ionized Helium $(He^+)$, $Z_{He} = 2$. The ratio of acceleration is $\frac{a_{He}}{a_H} = \frac{Z_{He}^3}{Z_H^3} = \frac{2^3}{1^3} = \frac{8}{1} = 8$.

The ratio of acceleration of the electron in a singly ionized Helium atom $(He^+)$ to that of a Hydrogen atom $(H)$ (both in the ground state) is:

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