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 $E_n{r}_n/E_1{r}_1$ as a function of $n$ is a straight line of slope $0$.
• B. plot of $r_n{v}_n/r_1{v}_1$ as a function of $n$ is a straight line of slope $1$.
• C. plot of $lnn\left(\frac{r_n}{r_1}\right)$ 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 $=4)$