MHT CET · Physics · Atomic Physics
In hydrogen atoms, radius of the smallest orbit of the electron is \(a_0\), the radius of the third orbit is
- A \(\frac{a_0}{9}\)
- B \(3 \mathrm{a}_0\)
- C \(6 a_0\)
- D \(9 \mathrm{a}_0\)
Answer & Solution
Correct Answer
(D) \(9 \mathrm{a}_0\)
Step-by-step Solution
Detailed explanation
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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