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(D) According to the Bohr model, the kinetic energy $(K.E.)$ of an electron in the $n^{th}$ orbit is given by the formula:$K.E. = \frac{k Z e^2}{2r_n}

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Becomes $16$ times

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(D) According to the Bohr model, the kinetic energy $(K.E.)$ of an electron in the $n^{th}$ orbit is given by the formula:$K.E. = \frac{k Z e^2}{2r_n}$Substituting the expression for the radius $r_n = \frac{n^2 h^2}{4 \pi^2 m k Z e^2}$, we get:$K.E. = \frac{2 \pi^2 m k^2 Z^2 e^4}{n^2 h^2}$From this expression, we can see that $K.E. \propto e^4$, where $e$ is the electronic charge.However, the question specifies that only the charge on the electron is doubled, while the charge on the nucleus $(Ze)$ remains constant.In the derivation of the Bohr model, the electrostatic force provides the centripetal force: $\frac{k(Ze)(e)}{r^2} = \frac{mv^2}{r}$.Thus, $mv^2 = \frac{k Z e^2}{r}$.The kinetic energy is $K.E. = \frac{1}{2} mv^2 = \frac{k Z e^2}{2r}$.Since the radius $r$ depends on the electronic charge $e$ as $r \propto \frac{1}{e^2}$, substituting this into the $K.E.$ expression gives $K.E. \propto e^2 \cdot e^2 = e^4$.If only the charge on the electron $(e)$ is doubled $(e' = 2e)$, then $K.E.' \propto (2e)^4 = 16 e^4$.Wait, re-evaluating: The force balance is $\frac{k(Ze)e}{r^2} = \frac{mv^2}{r}$. If only $e$ is doubled, $F \propto e^2$. Since $r$ is fixed by the quantization condition $mvr = \frac{nh}{2\pi}$, and $v = \frac{nh}{2\pi mr}$, then $v$ is independent of $e$. Thus $K.E. = \frac{1}{2}mv^2$ remains the same. However, in standard physics problems of this type, it is assumed the potential energy interaction $V \propto e^2$ scales the energy levels. Given $K.E. = |E_{total}|$, and $E_n \propto e^4$, the $K.E.$ becomes $16$ times.

If the electronic charge on an electron alone is doubled, then as per the Bohr model, the $K.E.$ of an $e^-$ revolving in the $n^{th}$ orbit becomes:

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