Question

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

• A. plot of $E_nr_n/E_1r_1$ as a function of $n$ is a straight line of slope $0$.
• B. plot of $r_nv_n/r_1v_1$ as a function of $n$ is a straight line of slope $1$.
• C. plot of $ln\, n \,\left(\frac{r_{n}}{r_{1}}\right)$ as a function of $ln(n)$ is a straight line of slope $2$.
• D. 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 $4$.

Answer 1

Answer: (A), (B), (C), (D)

We know that, $v \propto \frac{1}{n}$ (speed)
$E_{n} \propto \frac{1}{n^{2}} $ (energy)
$r_{n} \propto n^{2} $
$\therefore E_{n} r_{n} \propto n^{0} $
$\therefore E_{n} r_{n} \propto E_{1} r_{1}$
$ \frac{E_{n} r_{n}}{E_{1} r_{1}}=$ constant $(\because $ slope $=0) $
$ r_{n} v_{n} \propto n^{2} \times \frac{1}{n} \propto n$
$\therefore \frac{r_{n} v_{n}}{r_{1} v_{1}}=n (\because $ slope $=1) $
$ r_{n} \propto n^{2} $
$\therefore \frac{r_{n}}{r_{1}}=n^{2}$
$\log \left(\frac{r_{n}}{r_{1}}\right)=2 \log n (\because$ slope $=2) $
$\therefore \frac{r_{n}}{E_{n}} \propto n^{4} $
$ \frac{r_{n}}{E_{n}} \times \frac{E_{1}}{r_{1}}=n^{4}$
$ \log \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right)=4 \log (n)(\because $ slope $=4)$