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(A-D) According to Bohr's model for a hydrogen atom:$v_{n} \propto \frac{1}{n}$$E_{n} \propto \frac{1}{n^{2}}$$r_{n} \propto n^{2}$For option $A$: $\f

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plot of $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope $4$

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(A-D) According to Bohr's model for a hydrogen atom:$v_{n} \propto \frac{1}{n}$$E_{n} \propto \frac{1}{n^{2}}$$r_{n} \propto n^{2}$For option $A$: $\frac{E_{n} r_{n}}{E_{1} r_{1}} \propto \frac{(1/n^{2}) \cdot n^{2}}{1} = 1$. This is a constant, so the slope is $0$.For option $B$: $\frac{r_{n} v_{n}}{r_{1} v_{1}} \propto \frac{n^{2} \cdot (1/n)}{1} = n$. This is a straight line with slope $1$.For option $C$: $\frac{r_{n}}{r_{1}} = n^{2}$. Taking natural log on both sides: $\ln \left(\frac{r_{n}}{r_{1}}\right) = 2 \ln(n)$. This is a straight line with slope $2$.For option $D$: $\frac{r_{n}}{E_{n}} \propto \frac{n^{2}}{1/n^{2}} = n^{4}$. Thus, $\frac{r_{n} E_{1}}{E_{n} r_{1}} = n^{4}$. Taking natural log: $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right) = 4 \ln(n)$. This is a straight line with slope $4$.All options $A, B, C,$ and $D$ are mathematically correct based on Bohr's model.

Let $v_{n}$ and $E_{n}$ be the respective speed and energy of an electron in the $n$th orbit of radius $r_{n}$, in a hydrogen atom, as predicted by Bohr's model. Then:

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