Question

According to Bohr model of hydrogen atom, the radius of stationary orbit characterized by the principal quantum number n is proportional to

Answer 1

Step 1: Some points about the Bohr model

1. It is created by Neil and Rutherford in 1913.
2. The structure of the system is like a solar system.
3. An electron moves around the nucleus in a circular orbit.
4. It emits energy in the form of light.
5. It postulates that electrons orbit the nucleus at fixed energy levels.

Step 2: The formula of the Bohr model

Consider a hydrogen-like atom in which an electron of mass $m$ and charge $\left(-e\right)$ revolves in a circular orbit of radius $r$ with a velocity $v$ round a nucleus of charge $+Ze$. The electrostatic attractive force on the nucleus $\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$ provides the centripetal force $\frac{m{v}^2}{r}$ to keep the electron in orbit so that we can write

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{Z{e}^2}{{\mathrm{r}}^2}=$$\frac{m{v}^2}{r}$ which gives

$v^2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{Z{e}^2}{\mathrm{mr}}$

Here ${\epsilon }_0$ is the permittivity of vacuum and has the value ${10}^{-9}/36\mathrm{\pi }\mathrm{F}/\mathrm{m}$.

From Bohr's quantum condition, we have

$L=m{r}^2\omega =mvr=n\sout{h}$

where $\omega$ is the angular velocity and $v$ is the linear velocity of the electron in the orbit. From this equation we have

$v=\frac{n\sout{h}}{mr}$

Eliminating $v$ from both the equations we get

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{Z{e}^2}{{\mathrm{r}}^2}=$${\left(\frac{n\sout{h}}{mr}\right)}^2$ which gives

$r=\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{n}}^2{\sout{\mathrm{h}}}^2}{mZ{e}^2}$

Hence, we get

$v=\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{n}\sout{\mathrm{h}}}$

Thus both the radius of the orbit and the electron velocity depend on the quantum number $n$. The orbit $n=1$ has the smallest radius. For hydrogen $\left(Z=1\right)$, this radius is known as the Bohr radius and is given by

$a_0=\frac{4{\mathrm{\pi \epsilon }}_0{\sout{\mathrm{h}}}^2}{m{e}^2}=0.529\times {10}^{-10}m=0.529\stackrel{\circ }{A}$

where we have substituted the numerical values of ${\epsilon }_0,m,e$ and $\sout{h}$.

The radii of the orbits are directly proportional to $n^2$ so that the radii of the successive orbits are in the ratios $1:4:9:16:....$

$\therefore r\propto n^2$

Step 3: A diagram to represent the Bohr model

Therefore, now according to the Bohr model of the hydrogen atom, the radius of stationary orbit characterized by the principal quantum number n is proportional to $n^2$.