MHT CET · Physics · Atomic Physics
The ratio of the velocity of the electron in the first Bohr orbit to that in the second Bohr orbit of hydrogen atom is
- A \(8: 1\)
- B \(2: 1\)
- C \(4: 1\)
- D \(1: 4\)
Answer & Solution
Correct Answer
(B) \(2: 1\)
Step-by-step Solution
Detailed explanation
Velocity of electron in the \(\mathrm{n}^{\mathrm{th}}\) orbit \(\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{Ze}^2}{2 \varepsilon_0 \mathrm{nh}}\)
\(\begin{aligned}
& \Rightarrow \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\
& \quad \text { given } \mathrm{n}_1=1 \text {, and } \mathrm{n}_2=2, \\
& \therefore \quad \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{2}{1}
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\
& \quad \text { given } \mathrm{n}_1=1 \text {, and } \mathrm{n}_2=2, \\
& \therefore \quad \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{2}{1}
\end{aligned}\)
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