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In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to \(\frac{1}{3}\) rd of its initial value. The radius of the orbit in the ground state is \(\mathrm{R}\).
The radius of the orbit in that higher energy state is

A \(9 \mathrm{R}\)
B \(2 \mathrm{R}\)
C \(3 \mathrm{R}\)
D \(27 \mathrm{R}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(3 \mathrm{R}\)

Step-by-step Solution

Detailed explanation

According to Bohr's orbit
\(m v r=\frac{n h}{2 \pi}\)
i.e., \(\quad \mathrm{v} \propto \frac{1}{\mathrm{r}}, \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}}\)
\(\frac{v}{v / z}=\frac{r_{2}}{R} \Rightarrow r_{2}=3 R\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to \(\frac{1}{3}\) rd of its initial value. The radius of the orbit in the ground state is \(\mathrm{R}\).The radius of the orbit in that higher energy state is

Step-by-step Solution

In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to \(\frac{1}{3}\) rd of its initial value. The radius of the orbit in the ground state is \(\mathrm{R}\).
The radius of the orbit in that higher energy state is