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

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## Step 1: Some points about the Bohr modelIt is created by Neil and Rutherford in 1913.The structure of the system is like a solar system.An electron moves around the nucleus in a circular orbit.It emits energy in the form of light.It postulates that electrons orbit the nucleus at fixed energy levels.Step 2: The formula of the Bohr modelConsider 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}}=$$\frac{m{v}^{2}}{r}$ which gives${v}^{2}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}.\frac{Z{e}^{2}}{\mathrm{mr}}$Here ${\epsilon }_{0}$ is the permittivity of vacuum and has the value ${10}^{-9}/36\mathrm{\pi }\mathrm{F}/\mathrm{m}$.From Bohr's quantum condition, we have$L=m{r}^{2}\omega =mvr=n\overline{)h}$where $\omega$ is the angular velocity and $v$ is the linear velocity of the electron in the orbit. From this equation we have$v=\frac{n\overline{)h}}{mr}$Eliminating $v$ from both the equations we get$\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}.\frac{Z{e}^{2}}{{\mathrm{r}}^{2}}=$${\left(\frac{n\overline{)h}}{mr}\right)}^{2}$ which gives$r=\frac{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{n}}^{2}{\overline{)\mathrm{h}}}^{2}}{mZ{e}^{2}}$Hence, we get$v=\frac{Z{e}^{2}}{4{\mathrm{\pi \epsilon }}_{0}\mathrm{n}\overline{)\mathrm{h}}}$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${a}_{0}=\frac{4{\mathrm{\pi \epsilon }}_{0}{\overline{)\mathrm{h}}}^{2}}{m{e}^{2}}=0.529×{10}^{-10}m=0.529\stackrel{\circ }{A}$where we have substituted the numerical values of ${\epsilon }_{0},m,e$ and $\overline{)h}$.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 modelTherefore, 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 ${n}^{2}$.

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