MHT CET · Physics · Atomic Physics
In a hydrogen atom in its ground state, the first Bohr orbit has radius \(r_1\). The electron's orbital speed becomes one-third when the atom is raised to one of its excited states. The radius of the orbit in that excited state is
- A \(3 \mathrm{r}_1\)
- B \(4 r_1\)
- C \(9 \mathrm{r}_1\)
- D \(16 \mathrm{r}_1\)
Answer & Solution
Correct Answer
(C) \(9 \mathrm{r}_1\)
Step-by-step Solution
Detailed explanation
Velocity of electron
\(\begin{aligned}
& \mathrm{V}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\
& \frac{\mathrm{~V}_{\mathrm{n}}}{\mathrm{~V}_{\mathrm{n}}^{\prime}}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \\
& \frac{\mathrm{~V}_1}{\mathrm{~V}_1}=\frac{\mathrm{n}_2}{1} \\
& \therefore \quad \mathrm{n}_2=3 \\
& \operatorname{Radius~of~el~}^3 \\
& \mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^2 \\
& \frac{\mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{\prime}}=\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}\right)^2 \\
& \frac{r_1}{r_2}=\left(\frac{1}{3}\right)^2 \\
& r_2=9 r_1
\end{aligned}\)
Radius of electron
\(\begin{aligned}
& \mathrm{V}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\
& \frac{\mathrm{~V}_{\mathrm{n}}}{\mathrm{~V}_{\mathrm{n}}^{\prime}}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \\
& \frac{\mathrm{~V}_1}{\mathrm{~V}_1}=\frac{\mathrm{n}_2}{1} \\
& \therefore \quad \mathrm{n}_2=3 \\
& \operatorname{Radius~of~el~}^3 \\
& \mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^2 \\
& \frac{\mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{\prime}}=\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}\right)^2 \\
& \frac{r_1}{r_2}=\left(\frac{1}{3}\right)^2 \\
& r_2=9 r_1
\end{aligned}\)
Radius of electron
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