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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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plot of $\frac{E_{n} r_{n}}{E_{1} r_{1}}$ as a function of $n$ is a straight line of slope 0, plot of $\frac{r_{n} v_{n}}{r_{1} v_{1}}$ as a function of $n$ is a straight line of slope 1, plot of $\ln \left(\frac{r_{n}}{r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope 2, plot of In $\left(\frac{r E_{1}}{E_{n} r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope 4
We know that, $v \propto \frac{1}{n}$ (speed)
$E_{n} \propto \frac{1}{n^{2}} \quad(\text { energy })$
$r_{n} \propto n^{2}$
$\therefore E_{n T_{n}} \propto n^{0}$
$\therefore E_{n} r_{n} \propto E_{1} r_{1}$
$\frac{E_{1} r_{1}}{E_{1} r_{1}}$
$r_{n} v_{s}=n^{2} \times \frac{1}{n} \propto n = constant$
$\therefore \frac{r_{n} v_{n}}{r_{1} v_{1}}=n$
$r_{\mathrm{e}} \sim n^{2}$
$\therefore \frac{r_{\mathrm{n}}}{r_{1}}=n^{2}$
$\log \left(\frac{r_{\mathrm{n}}}{r_{1}}\right)=2 \log n \quad(\because$ slope $=2)$
$\frac{r_{n}}{E_{n}}=n^{4}$
$\therefore \frac{r_{n}}{E_{n}} \times \frac{E_{1}}{r_{1}}=n^{4}$
$\log \left(\frac{r_{1} E_{1}}{E_{\alpha} r_{1}}\right)=4 \log (n)(\because$ slope $=4)$
$E_{n} \propto \frac{1}{n^{2}} \quad(\text { energy })$
$r_{n} \propto n^{2}$
$\therefore E_{n T_{n}} \propto n^{0}$
$\therefore E_{n} r_{n} \propto E_{1} r_{1}$
$\frac{E_{1} r_{1}}{E_{1} r_{1}}$
$r_{n} v_{s}=n^{2} \times \frac{1}{n} \propto n = constant$
$\therefore \frac{r_{n} v_{n}}{r_{1} v_{1}}=n$
$r_{\mathrm{e}} \sim n^{2}$
$\therefore \frac{r_{\mathrm{n}}}{r_{1}}=n^{2}$
$\log \left(\frac{r_{\mathrm{n}}}{r_{1}}\right)=2 \log n \quad(\because$ slope $=2)$
$\frac{r_{n}}{E_{n}}=n^{4}$
$\therefore \frac{r_{n}}{E_{n}} \times \frac{E_{1}}{r_{1}}=n^{4}$
$\log \left(\frac{r_{1} E_{1}}{E_{\alpha} r_{1}}\right)=4 \log (n)(\because$ slope $=4)$
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