In Bohr's atomic model,the radius (r) and ene | vedclass

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Question

(A) In Bohr's atomic model,the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,which implies $r \propto n^2$.

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$E \propto -1/n^2, r \propto n^2$

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(A) In Bohr's atomic model,the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,which implies $r \propto n^2$. The total energy of an electron in the $n^{th}$ orbit is $E_n = -\frac{m Z^2 e^4}{8 \epsilon_0^2 h^2 n^2}$,which implies $E \propto -1/n^2$. Considering the magnitude of energy,the relationship is $r \propto n^2$ and $E \propto 1/n^2$.

In Bohr's atomic model,the radius $(r)$ and energy $(E)$ of an orbit $(n)$ are related as:

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