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In hydrogen atoms, radius of the smallest orbit of the electron is $a_0$, the radius of the third orbit is
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The correct answer is:
$9 \mathrm{a}_0$
Concept: Bohr orbit radius
The electron undergoes a uniform circular motion in Bohr's orbit. The coulombic attraction of the nucleus balancing the centrifugal force:
$\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{eZ}}{4 \pi \varepsilon_0 \mathrm{r}^2}$
Further, the angular momentum of electron in Bohr's orbit is:
$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$
On squaring the second equation and dividing it by the first equation:
$r \propto \frac{n^2}{Z}$
The smallest Bohr orbit has radius $\mathrm{a}_0$ as $\mathrm{n}=1$. Using the above functional dependence of the Bohr orbit, the third orbit is has radius $9 \mathrm{a}_0$.
The electron undergoes a uniform circular motion in Bohr's orbit. The coulombic attraction of the nucleus balancing the centrifugal force:
$\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{eZ}}{4 \pi \varepsilon_0 \mathrm{r}^2}$
Further, the angular momentum of electron in Bohr's orbit is:
$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$
On squaring the second equation and dividing it by the first equation:
$r \propto \frac{n^2}{Z}$
The smallest Bohr orbit has radius $\mathrm{a}_0$ as $\mathrm{n}=1$. Using the above functional dependence of the Bohr orbit, the third orbit is has radius $9 \mathrm{a}_0$.
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